In a \u2206ABC, if \u2220A = 55\u00b0, \u2220B = 40\u00b0, find \u2220C.
Solution:
<\/strong>\u2235 Sum of three angles of a triangle is 180\u00b0
\u2234 In \u2206ABC, \u2220A = 55\u00b0, \u2220B = 40\u00b0
But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)
<\/p>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040207_337b02451b_o-1.png\")
<\/figure>\n\n\n\n
\u21d2 55\u00b0 + 40\u00b0 + \u2220C = 180\u00b0
\u21d2 95\u00b0 + \u2220C = 180\u00b0
\u2234 \u2220C= 180\u00b0 -95\u00b0 = 85\u00b0<\/p>\n\n\n\n
Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Mark the correct alternative in each of the following: Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Question 25.
If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
<\/strong>Ratio in three angles of a triangle =1:2:3
Let first angle = x
Then second angle = 2x
and third angle = 3x
\u2234 x + 2x + 3x = 180\u00b0 (Sum of angles of a triangle)
\u21d26x = 180\u00b0
\u21d2x = 180\u22186 = 30\u00b0
\u2234 First angle = x = 30\u00b0
Second angle = 2x = 2 x 30\u00b0 = 60\u00b0
and third angle = 3x = 3 x 30\u00b0 = 90\u00b0
\u2234 Angles are 30\u00b0, 60\u00b0, 90\u00b0<\/p>\n\n\n\n
The angles of a triangle are (x \u2013 40)\u00b0, (x \u2013 20)\u00b0 and (12 x \u2013 10)\u00b0. Find the value of x.
Solution:
<\/strong>\u2235 Sum of three angles of a triangle = 180\u00b0
\u2234 (x \u2013 40)\u00b0 + (x \u2013 20)\u00b0 + (12x-10)0 = 180\u00b0
\u21d2 x \u2013 40\u00b0 + x \u2013 20\u00b0 + 12x \u2013 10\u00b0 = 180\u00b0
\u21d2 x + x+ 12x \u2013 70\u00b0 = 180\u00b0
\u21d2 52x = 180\u00b0 + 70\u00b0 = 250\u00b0
\u21d2 x = 250\u2218x25 = 100\u00b0
\u2234 x = 100\u00b0<\/p>\n\n\n\n
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30\u00b0. Determine all the angles of the triangle.
Solution:
<\/strong>Let each of the two equal angles = x
Then third angle = x + 30\u00b0
But sum of the three angles of a triangle is 180\u00b0
\u2234 x + x + x + 30\u00b0 = 180\u00b0
\u21d2 3x + 30\u00b0 = 180\u00b0
\u21d23x = 150\u00b0 \u21d2x = 150\u22183 = 50\u00b0
\u2234 Each equal angle = 50\u00b0
and third angle = 50\u00b0 + 30\u00b0 = 80\u00b0
\u2234 Angles are 50\u00b0, 50\u00b0 and 80\u00b0<\/p>\n\n\n\n
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
<\/strong>In the triangle ABC,
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040037_61825e9d82_o.png\")
<\/figure>\n\n\n\n
\u2220B = \u2220A + \u2220C
But \u2220A + \u2220B + \u2220C = 180\u00b0
\u21d2\u2220B + \u2220A + \u2220C = 180\u00b0
\u21d2\u2220B + \u2220B = 180\u00b0
\u21d22\u2220B = 180\u00b0
\u2234 \u2220B = 180\u22182 = 90\u00b0
\u2235 One angle of the triangle is 90\u00b0
\u2234 \u2206ABC is a right triangle.<\/p>\n\n\n\n
Can a triangle have:
(i) Two right angles?
(ii) Two obtuse angles?
(iii) Two acute angles?
(iv) All angles more than 60\u00b0?
(v) All angles less than 60\u00b0?
(vi) All angles equal to 60\u00b0?
Justify your answer in each case.
Solution:
<\/strong>(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180\u00b0 and if there are two right-angles, then the third angle will be zero which is not possible.
(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180\u00b0 and if there are
two obtuse angle, then the third angle will be negative which is not possible.
(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180\u00b0.
(iv) All angles more than 60\u00b0, they are also not possible as the sum will be more than 180\u00b0.
(v) All angles less than 60\u00b0. They are also not possible as the sum will be less than 180\u00b0.
(vi) All angles equal to 60\u00b0. This is possible as the sum will be 60\u00b0 x 3 = 180\u00b0.<\/p>\n\n\n\n
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10\u00b0, find the three angles.
Solution:<\/strong>
Let three angles of a triangle be x\u00b0, (x + 10)\u00b0, (x + 20)\u00b0
But sum of three angles of a triangle is 180\u00b0
\u2234 x + (x+ 10)\u00b0 + (x + 20) = 180\u00b0
\u21d2 x + x+10\u00b0+ x + 20 = 180\u00b0
\u21d2 3x + 30\u00b0 = 180\u00b0
\u21d2 3x = 180\u00b0 \u2013 30\u00b0 = 150\u00b0
\u2234 x = 180\u22182 = 50\u00b0
\u2234 Angle are 50\u00b0, 50 + 10, 50 + 20
i.e. 50\u00b0, 60\u00b0, 70\u00b0<\/p>\n\n\n\n
ABC is a triangle is which \u2220A = 72\u00b0, the internal bisectors of angles B and C meet in O. Find the magnitude of \u2220BOC.
Solution:<\/strong>
In \u2206ABC, \u2220A = 12\u00b0 and bisectors of \u2220B and \u2220C meet at O
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703039897_69de4ee965_o.png\")
<\/figure>\n\n\n\n
Now \u2220B + \u2220C = 180\u00b0 \u2013 12\u00b0 = 108\u00b0
\u2235 OB and OC are the bisectors of \u2220B and \u2220C respectively
\u2234 \u2220OBC + \u2220OCB = 12 (B + C)
= 12 x 108\u00b0 = 54\u00b0
But in \u2206OBC,
\u2234 \u2220OBC + \u2220OCB + \u2220BOC = 180\u00b0
\u21d2 54\u00b0 + \u2220BOC = 180\u00b0
\u2220BOC = 180\u00b0-54\u00b0= 126\u00b0
OR
According to corollary,
\u2220BOC = 90\u00b0+ 12 \u2220A
= 90+ 12 x 72\u00b0 = 90\u00b0 + 36\u00b0 = 126\u00b0<\/p>\n\n\n\n
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Solution:<\/strong>
In right \u2206ABC, \u2220A is the vertex angle and OB and OC are the bisectors of \u2220B and \u2220C respectively
To prove : \u2220BOC cannot be a right angle
Proof: \u2235 OB and OC are the bisectors of \u2220B and \u2220C respectively
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703039697_361ceff1b1_o.png\")
<\/figure>\n\n\n\n
\u2234 \u2220BOC = 90\u00b0 x 12 \u2220A
Let \u2220BOC = 90\u00b0, then
12 \u2220A = O
\u21d2\u2220A = O
Which is not possible because the points A, B and C will be on the same line Hence, \u2220BOC cannot be a right angle.<\/p>\n\n\n\n
If the bisectors of the base angles of a triangle enclose an angle of 135\u00b0. Prove that the triangle is a right triangle.
Solution:<\/strong>
Given : In \u2206ABC, OB and OC are the bisectors of \u2220B and \u2220C and \u2220BOC = 135\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703039507_df9d86438a_o.png\")
<\/figure>\n\n\n\n
To prove : \u2206ABC is a right angled triangle
Proof: \u2235 Bisectors of base angles \u2220B and \u2220C of the \u2206ABC meet at O
\u2234 \u2220BOC = 90\u00b0+ 12\u2220A
But \u2220BOC =135\u00b0
\u2234 90\u00b0+ 12 \u2220A = 135\u00b0
\u21d2 12\u2220A= 135\u00b0 -90\u00b0 = 45\u00b0
\u2234 \u2220A = 45\u00b0 x 2 = 90\u00b0
\u2234 \u2206ABC is a right angled triangle<\/p>\n\n\n\n
In a \u2206ABC, \u2220ABC = \u2220ACB and the bisectors of \u2220ABC and \u2220ACB intersect at O such that \u2220BOC = 120\u00b0. Show that \u2220A = \u2220B = \u2220C = 60\u00b0.
Solution:<\/strong>
Given : In \u2220ABC, BO and CO are the bisectors of \u2220B and \u2220C respectively and \u2220BOC = 120\u00b0 and \u2220ABC = \u2220ACB
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703039197_da3a724242_o.png\")
<\/figure>\n\n\n\n
To prove : \u2220A = \u2220B = \u2220C = 60\u00b0
Proof : \u2235 BO and CO are the bisectors of \u2220B and \u2220C
\u2234 \u2220BOC = 90\u00b0 + 12\u2220A
But \u2220BOC = 120\u00b0
\u2234 90\u00b0+ 12 \u2220A = 120\u00b0
\u2234 12 \u2220A = 120\u00b0 \u2013 90\u00b0 = 30\u00b0
\u2234 \u2220A = 60\u00b0
\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Angles of a triangle)
\u2220B + \u2220C = 180\u00b0 \u2013 60\u00b0 = 120\u00b0 and \u2220B = \u2220C
\u2235 \u2220B = \u2220C = 120\u22182 = 60\u00b0
Hence \u2220A = \u2220B = \u2220C = 60\u00b0<\/p>\n\n\n\n
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:<\/strong>
In a \u2206ABC,
Let \u2220A < \u2220B + \u2220C
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593358422_f37d88c7cf_o.png\")
<\/figure>\n\n\n\n
\u21d2\u2220A + \u2220A < \u2220A + \u2220B + \u2220C
\u21d2 2\u2220A < 180\u00b0
\u21d2 \u2220A < 90\u00b0 (\u2235 Sum of angles of a triangle is 180\u00b0)
Similarly, we can prove that
\u2220B < 90\u00b0 and \u2220C < 90\u00b0
\u2234 Each angle of the triangle are acute angle.<\/p>\n\n\n\nEx 11.2<\/h4>\n\n\n\n
The exterior angles obtained on producing the base of a triangle both ways are 104\u00b0 and 136\u00b0. Find all the angles of the triangle.
Solution:<\/strong>
In \u2206ABC, base BC is produced both ways to D and E respectivley forming \u2220ABE = 104\u00b0 and \u2220ACD = 136\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703039077_e49a07397c_o.png\")
<\/figure>\n\n\n\n
In the figure, the sides BC, CA and AB of a \u2206ABC have been produced to D, E and F respectively. If \u2220ACD = 105\u00b0 and \u2220EAF = 45\u00b0, find all the angles of the \u2206ABC.
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593358102_a3d2c61d79_o.png\")
<\/figure>\n\n\n\n
Solution<\/strong>:
In \u2206ABC, sides BC, CA and BA are produced to D, E and F respectively.
\u2220ACD = 105\u00b0 and \u2220EAF = 45\u00b0
\u2220ACD + \u2220ACB = 180\u00b0 (Linear pair)
\u21d2 105\u00b0 + \u2220ACB = 180\u00b0
\u21d2 \u2220ACB = 180\u00b0- 105\u00b0 = 75\u00b0
\u2220BAC = \u2220EAF (Vertically opposite angles)
= 45\u00b0
But \u2220BAC + \u2220ABC + \u2220ACB = 180\u00b0
\u21d2 45\u00b0 + \u2220ABC + 75\u00b0 = 180\u00b0
\u21d2 120\u00b0 +\u2220ABC = 180\u00b0
\u21d2 \u2220ABC = 180\u00b0- 120\u00b0
\u2234 \u2220ABC = 60\u00b0
Hence \u2220ABC = 60\u00b0, \u2220BCA = 75\u00b0
and \u2220BAC = 45\u00b0<\/p>\n\n\n\n
Compute the value of x in each of the following figures:
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703038737_b28e3e2e33_o.png\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593357912_c8ecf52f5f_o.png\")
<\/figure>\n\n\n\n
Solution<\/strong>:
(i) In \u2206ABC, sides BC and CA are produced to D and E respectively
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643870021_5b1428fdbb_o.png\")
<\/figure>\n\n\n\n
(ii) In \u2206ABC, side BC is produced to either side to D and E respectively
\u2220ABE = 120\u00b0 and \u2220ACD =110\u00b0
\u2235 \u2220ABE + \u2220ABC = 180\u00b0 (Linear pair)
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593357622_0e671134b5_o.png\")
<\/figure>\n\n\n\n
(iii) In the figure, BA || DC
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643869581_4277c744cb_o.png\")
<\/figure>\n\n\n\n
In the figure, AC \u22a5 CE and \u2220A: \u2220B : \u2220C = 3:2:1, find the value of \u2220ECD.<\/strong>
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643869171_a057c46844_o.png\")
<\/figure>\n\n\n\n
Solution:
<\/strong>In \u2206ABC, \u2220A : \u2220B : \u2220C = 3 : 2 : 1
BC is produced to D and CE \u22a5 AC
\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangles)
Let\u2220A = 3x, then \u2220B = 2x and \u2220C = x
\u2234 3x + 2x + x = 180\u00b0 \u21d2 6x = 180\u00b0
\u21d2 x = 180\u22186 = 30\u00b0
\u2234 \u2220A = 3x = 3 x 30\u00b0 = 90\u00b0
\u2220B = 2x = 2 x 30\u00b0 = 60\u00b0
\u2220C = x = 30\u00b0
In \u2206ABC,
Ext. \u2220ACD = \u2220A + \u2220B
\u21d2 90\u00b0 + \u2220ECD = 90\u00b0 + 60\u00b0 = 150\u00b0
\u2234 \u2220ECD = 150\u00b0-90\u00b0 = 60\u00b0<\/p>\n\n\n\n
In the figure, AB || DE, find \u2220ACD.<\/strong>
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593357032_3d6c2881aa_o.png\")
<\/figure>\n\n\n\n
Solution:
<\/strong>In the figure, AB || DE
AE and BD intersect each other at C \u2220BAC = 30\u00b0 and \u2220CDE = 40\u00b0
\u2235 AB || DE
\u2234 \u2220ABC = \u2220CDE (Alternate angles)
<\/p>\n\n\n\n\n\n
\u21d2 \u2220ABC = 40\u00b0
In \u2206ABC, BC is produced
Ext. \u2220ACD = Int. \u2220A + \u2220B
= 30\u00b0 + 40\u00b0 = 70\u00b0<\/p>\n\n\n\n
Which of the following statements are true (T) and which are false (F):
(i) Sum of the three angles of a triangle is 180\u00b0.
(ii) A triangle can have two right angles.
(iii) All the angles of a triangle can be less than 60\u00b0.
(iv) All the angles of a triangle can be greater than 60\u00b0.
(v) All the angles of a triangle can be equal to 60\u00b0.
(vi) A triangle can have two obtuse angles.
(vii) A triangle can have at most one obtuse angles.
(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.
(ix) An exterior angle of a triangle is less than either of its interior opposite angles.
(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
(xi) An exterior angle of a triangle is greater than the opposite interior angles.
Solution:<\/strong>
(i) True.
(ii) False. A right triangle has only one right angle.
(iii) False. In this, the sum of three angles will be less than 180\u00b0 which is not true.
(iv) False. In this, the sum of three angles will be more than 180\u00b0 which is not true.
(v) True. As sum of three angles will be 180\u00b0 which is true.
(vi) False. A triangle has only one obtuse angle.
(vii) True.
(viii)True.
(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.
(x) True.
(xi) True.<\/p>\n\n\n\n
Fill in the blanks to make the following statements true:
(i) Sum of the angles of a triangle is \u2026\u2026\u2026
(ii) An exterior angle of a triangle is equal to the two \u2026\u2026.. opposite angles.
(iii) An exterior angle of a triangle is always \u2026\u2026.. than either of the interior opposite angles.
(iv) A triangle cannot have more than \u2026\u2026\u2026. right angles.
(v) A triangles cannot have more than \u2026\u2026\u2026 obtuse angles.
Solution:<\/strong>
(i) Sum of the angles of a triangle is 180\u00b0.
(ii) An exterior angle of a triangle is equal to the two interior opposite angles.
(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.
(iv) A triangle cannot have more than one right angles.
(v) A triangles cannot have more than one obtuse angles.<\/p>\n\n\n\n
In a \u2206ABC, the internal bisectors of \u2220B and \u2220C meet at P and the external bisectors of \u2220B and \u2220C meet at Q. Prove that \u2220BPC + \u2220BQC = 180\u00b0.
Solution:<\/strong>
Given : In \u2206ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior \u2220B and \u2220C meet at P and bisectors of exterior angles B and C meet at Q.
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593356782_846e6279f7_o.png\")
<\/figure>\n\n\n\n
To prove : \u2220BPC + \u2220BQC = 180\u00b0
Proof : \u2235 PB and PC are the internal bisectors of \u2220B and \u2220C
\u2220BPC = 90\u00b0+ 12 \u2220A \u2026(i)
Similarly, QB and QC are the bisectors of exterior angles B and C
\u2234 \u2220BQC = 90\u00b0 + 12 \u2220A \u2026(ii)
Adding (i) and (ii),
\u2220BPC + \u2220BQC = 90\u00b0 + 12 \u2220A + 90\u00b0 \u2013 12 \u2220A
= 90\u00b0 + 90\u00b0 = 180\u00b0
Hence \u2220BPC + \u2220BQC = 180\u00b0<\/p>\n\n\n\n
In the figure, compute the value of x.<\/strong>
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593356622_a26904c9aa_o.png\")
<\/figure>\n\n\n\n
Solution:
<\/strong>In the figure,
\u2220ABC = 45\u00b0, \u2220BAD = 35\u00b0 and \u2220BCD = 50\u00b0 Join BD and produce it E
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643868601_2fd5a4eac4_o.png\")
<\/figure>\n\n\n\n
In the figure, AB divides \u2220D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.<\/strong>
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643868361_ff6f3b98d0_o.png\")
<\/figure>\n\n\n\n
Solution:
<\/strong>In the figure AB = DB
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593356382_df4f219f84_o.png\")
<\/figure>\n\n\n\n
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703037367_3fb8dd1e90_o.png\")
<\/figure>\n\n\n\n
ABC is a triangle. The bisector of the exterior angle at B and the bisector of \u2220C intersect each other at D. Prove that \u2220D = 12 \u2220A.
Solution:<\/strong>
Given : In \u2220ABC, CB is produced to E bisectors of ext. \u2220ABE and into \u2220ACB meet at D.
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593356172_40bff58f4c_o.png\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643868071_8f2185882d_o.png\")
<\/figure>\n\n\n\n
In the figure, AM \u22a5 BC and AN is the bisector of \u2220A. If \u2220B = 65\u00b0 and \u2220C = 33\u00b0, find \u2220MAN.
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593355992_8103016355_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703037067_fb84003d03_o.png\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593355872_1d95c9f5bf_o.png\")
<\/figure>\n\n\n\n
In a AABC, AD bisects \u2220A and \u2220C > \u2220B. Prove that \u2220ADB > \u2220ADC.
Solution:<\/strong>
Given : In \u2206ABC,
\u2220C > \u2220B and AD is the bisector of \u2220A
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593355462_3a93a0243d_o.png\")
<\/figure>\n\n\n\n
To prove : \u2220ADB > \u2220ADC
Proof: In \u2206ABC, AD is the bisector of \u2220A
\u2234 \u22201 = \u22202
In \u2206ADC,
Ext. \u2220ADB = \u2220l+ \u2220C
\u21d2 \u2220C = \u2220ADB \u2013 \u22201 \u2026(i)
Similarly, in \u2206ABD,
Ext. \u2220ADC = \u22202 + \u2220B
\u21d2 \u2220B = \u2220ADC \u2013 \u22202 \u2026(ii)
From (i) and (ii)
\u2235 \u2220C > \u2220B (Given)
\u2234 (\u2220ADB \u2013 \u22201) > (\u2220ADC \u2013 \u22202)
But \u22201 = \u22202
\u2234 \u2220ADB > \u2220ADC<\/p>\n\n\n\n
In \u2206ABC, BD \u22a5 AC and CE \u22a5 AB. If BD and CE intersect at O, prove that \u2220BOC = 180\u00b0-\u2220A.
Solution:<\/strong>
Given : In \u2206ABC, BD \u22a5 AC and CE\u22a5 AB BD and CE intersect each other at O
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593355192_62f1936a30_o.png\")
<\/figure>\n\n\n\n
To prove : \u2220BOC = 180\u00b0 \u2013 \u2220A
Proof: In quadrilateral ADOE
\u2220A + \u2220D + \u2220DOE + \u2220E = 360\u00b0 (Sum of angles of quadrilateral)
\u21d2 \u2220A + 90\u00b0 + \u2220DOE + 90\u00b0 = 360\u00b0
\u2220A + \u2220DOE = 360\u00b0 \u2013 90\u00b0 \u2013 90\u00b0 = 180\u00b0
But \u2220BOC = \u2220DOE (Vertically opposite angles)
\u21d2 \u2220A + \u2220BOC = 180\u00b0
\u2234 \u2220BOC = 180\u00b0 \u2013 \u2220A<\/p>\n\n\n\n
In the figure, AE bisects \u2220CAD and \u2220B = \u2220C. Prove that AE || BC.
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593354882_d870a95ebe_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In AABC, BA is produced and AE is the bisector of \u2220CAD
\u2220B = \u2220C
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593354722_ae04496774_o.png\")
<\/figure>\n\n\n\n
To prove : AE || BC
Proof: In \u2206ABC, BA is produced
\u2234 Ext. \u2220CAD = \u2220B + \u2220C
\u21d2 2\u2220EAC = \u2220C + \u2220C (\u2235 AE is the bisector of \u2220CAE) (\u2235 \u2220B = \u2220C)
\u21d2 2\u2220EAC = 2\u2220C
\u21d2 \u2220EAC = \u2220C
But there are alternate angles
\u2234 AE || BC<\/p>\n\n\n\nCoordinate Geometry VSAQS<\/h4>\n\n\n\n
Define a triangle.
Solution:<\/strong>
A figure bounded by three lines segments in a plane is called a triangle.<\/p>\n\n\n\n
Write the sum of the angles of an obtuse triangle.
Solution:<\/strong>
The sum of angles of an obtuse triangle is 180\u00b0.<\/p>\n\n\n\n
In \u2206ABC, if \u2220B = 60\u00b0, \u2220C = 80\u00b0 and the bisectors of angles \u2220ABC and \u2220ACB meet at a point O, then find the measure of \u2220BOC.
Solution:
In \u2206ABC, \u2220B = 60\u00b0, \u2220C = 80\u00b0
OB and OC are the bisectors of \u2220B and \u2220C
\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)
\u21d2 \u2220A + 60\u00b0 + 80\u00b0 = 180\u00b0
\u21d2 \u2220A + 140\u00b0 = 180\u00b0
\u2234 \u2220A = 180\u00b0- 140\u00b0 = 40\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771518608_97b01672cf_o.png\")
<\/figure>\n\n\n\n
= 90\u00b0 + \u2013 x 40\u00b0 = 90\u00b0 + 20\u00b0 = 110\u00b0<\/p>\n\n\n\n
If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.
Solution:
Sum of angles of a triangle = 180\u00b0
Ratio in the angles = 2 : 1 : 3
Let first angle = 2x
Second angle = x
and third angle = 3x
\u2234 2x + x + 3x = 180\u00b0 \u21d2 6x = 180\u00b0
\u2234 x = 180\u22186 = 30\u00b0
\u2234 First angle = 2x = 2 x 30\u00b0 = 60\u00b0
Second angle = x = 30\u00b0
and third angle = 3x = 3 x 30\u00b0 = 90\u00b0
Hence angles are 60\u00b0, 30\u00b0, 90\u00b0<\/p>\n\n\n\n
State exterior angle theorem.
Solution:<\/strong>
Given : In \u2206ABC, side BC is produced to D
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45643873851_bd99b68b47_o.png\")
<\/figure>\n\n\n\n
To prove : \u2220ACD = \u2220A + \u2220B
Proof: In \u2206ABC,
\u2220A + \u2220B + \u2220ACB = 180\u00b0 \u2026(i) (Sum of angles of a triangle)
and \u2220ACD + \u2220ACB = 180\u00b0 \u2026(ii) (Linear pair)
From (i) and (ii)
\u2220ACD + \u2220ACB = \u2220A + \u2220B + \u2220ACB
\u2220ACD = \u2220A + \u2220B
Hence proved.<\/p>\n\n\n\n
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
Solution:<\/strong>
In \u2206ABC,
\u2220A + \u2220C = \u2220B
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593360382_0f99b8b496_o.png\")
<\/figure>\n\n\n\n
But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)
\u2234 \u2220B + \u2220A + \u2220C = 180\u00b0
\u21d2 \u2220B + \u2220B = 180\u00b0
\u21d2 2\u2220B = 180\u00b0
\u21d2 \u2220B = 180\u22182 = 90\u00b0
\u2234 Third angle = 90\u00b0<\/p>\n\n\n\n
In the figure, if AB || CD, EF || BC, \u2220BAC = 65\u00b0 and \u2220DHF = 35\u00b0, find \u2220AGH.
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771518268_c234373109_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In figure, AB || CD, EF || BC \u2220BAC = 65\u00b0, \u2220DHF = 35\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593360252_0f22b72aac_o.png\")
<\/figure>\n\n\n\n
\u2235 EF || BC
\u2234 \u2220A = \u2220ACH (Alternate angle)
\u2234 \u2220ACH = 65\u00b0
\u2235\u2220GHC = \u2220DHF
(Vertically opposite angles)
\u2234 \u2220GHC = 35\u00b0
Now in \u2206GCH,
Ext. \u2220AGH = \u2220GCH + \u2220GHC
= 65\u00b0 + 35\u00b0 = 100\u00b0<\/p>\n\n\n\n
In the figure, if AB || DE and BD || FG such that \u2220FGH = 125\u00b0 and \u2220B = 55\u00b0, find x and y.
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593360182_d55798124c_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, AB || DF, BD || FG
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040957_cd16679562_o.png\")
<\/figure>\n\n\n\n
\u2220FGH = 125\u00b0 and \u2220B = 55\u00b0
\u2220FGH + FGE = 180\u00b0 (Linear pair)
\u21d2 125\u00b0 + y \u2013 180\u00b0
\u21d2 y= 180\u00b0- 125\u00b0 = 55\u00b0
\u2235 BA || FD and BD || FG
\u2220B = \u2220F = 55\u00b0
Now in \u2206EFG,
\u2220F + \u2220FEG + \u2220FGE = 180\u00b0
(Angles of a triangle)
\u21d2 55\u00b0 + x + 55\u00b0 = 180\u00b0
\u21d2 x+ 110\u00b0= 180\u00b0
\u2234 x= 180\u00b0- 110\u00b0 = 70\u00b0
Hence x = 70, y = 55\u00b0<\/p>\n\n\n\n
If the angles A, B and C of \u2206ABC satisfy the relation B \u2013 A = C \u2013 B, then find the measure of \u2220B.
Solution:<\/strong>
In \u2206ABC,
\u2220A + \u2220B + \u2220C= 180\u00b0 \u2026(i)
and B \u2013 A = C \u2013 B
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593360002_b52f470a73_o.png\")
<\/figure>\n\n\n\n
\u21d2 B + B = A + C \u21d2 2B = A + C
From (i),
B + 2B = 180\u00b0 \u21d2 3B = 180\u00b0
\u2220B = 180\u22183 = 60\u00b0
Hence \u2220B = 60\u00b0<\/p>\n\n\n\n
In \u2206ABC, if bisectors of \u2220ABC and \u2220ACB intersect at O at angle of 120\u00b0, then find the measure of \u2220A.
Solution:<\/strong>
In \u2206ABC, bisectors of \u2220B and \u2220C intersect at O and \u2220BOC = 120\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040777_27f07730c5_o.png\")
<\/figure>\n\n\n\n
But \u2220BOC = 90\u00b0+ 12
90\u00b0+ 12 \u2220A= 120\u00b0
\u21d2 12 \u2220A= 120\u00b0-90\u00b0 = 30\u00b0
\u2234 \u2220A = 2 x 30\u00b0 = 60\u00b0<\/p>\n\n\n\n
If the side BC of \u2206ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and \u2220A.
Solution:<\/strong>
In \u2206ABC, side BC is produced on both sides forming exterior \u2220ABE and \u2220ACD
Ext. \u2220ABE = \u2220A + \u2220ACB
and Ext. \u2220ACD = \u2220ABC + \u2220A
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040647_79722b213b_o.png\")
<\/figure>\n\n\n\n
Adding we get,
\u2220ABE + \u2220ACD = \u2220A + \u2220ACB + \u2220A + \u2220ABC
\u21d2 \u2220ABE + \u2220ACD \u2013 \u2220A = \u2220A 4- \u2220ACB + \u2220A + \u2220ABC \u2013 \u2220A (Subtracting \u2220A from both sides)
= \u2220A + \u2220ABC + \u2220ACB = \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<\/p>\n\n\n\n
In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find \u2220ACD: \u2220ADC.
Solution:<\/strong>
In \u2206ABC, AB = AC
AB is produced to D such that BD = BC
DC are joined
In \u2206ABC, AB = AC
\u2234 \u2220ABC = \u2220ACB
In \u2206 BCD, BD = BC
\u2234 \u2220BDC = \u2220BCD
and Ext. \u2220ABC = \u2220BDC + \u2220BCD = 2\u2220BDC (\u2235 \u2220BDC = \u2220BCD)
\u21d2 \u2220ACB = 2\u2220BCD (\u2235 \u2220ABC = \u2220ACB)
Adding \u2220BDC to both sides
\u21d2 \u2220ACB + \u2220BDC = 2\u2220BDC + \u2220BDC
\u21d2 \u2220ACB + \u2220BCD = 3 \u2220BDC (\u2235 \u2220BDC = \u2220BCD)
\u21d2 \u2220ACB = 3\u2220BDC<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040537_22366857b0_o.png\")
<\/figure>\n\n\n\n
In the figure, side BC of AABC is produced to point D such that bisectors of \u2220ABC and \u2220ACD meet at a point E. If \u2220BAC = 68\u00b0, find \u2220BEC.
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/45593359702_c97068d33b_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure,
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/30703040377_07031df438_o.png\")
<\/figure>\n\n\n\n
side BC of \u2206ABC is produced to D such that bisectors of \u2220ABC and \u2220ACD meet at E
\u2220BAC = 68\u00b0
In \u2206ABC,
Ext. \u2220ACD = \u2220A + \u2220B
\u21d2 12 \u2220ACD = 12 \u2220A + 12 \u2220B
\u21d2 \u22202= 12 \u2220A + \u22201 \u2026(i)
But in \u2206BCE,
Ext. \u22202 = \u2220E + \u2220l
\u21d2 \u2220E + \u2220l = \u22202 = 12 \u2220A + \u2220l [From (i)]
\u21d2 \u2220E = 12 \u2220A = 68\u22182 =34\u00b0<\/p>\n\n\n\nCoordinate Geometry MCQS<\/h4>\n\n\n\n
Question 1.
If all the three angles of a triangle are equal, then each one of them is equal to
(a) 90\u00b0
(b) 45\u00b0
(c) 60\u00b0
(d) 30\u00b0
Solution:<\/strong>
\u2235 Sum of three angles of a triangle = 180\u00b0
\u2234 Each angle = 180\u22183 = 60\u00b0 (c)<\/p>\n\n\n\n
If two acute angles of a right triangle are equal, then each acute is equal to
(a) 30\u00b0
(b) 45\u00b0
(c) 60\u00b0
(d) 90\u00b0
Solution:<\/strong>
In a right triangle, one angle = 90\u00b0
\u2234 Sum of other two acute angles = 180\u00b0 \u2013 90\u00b0 = 90\u00b0
\u2235 Both angles are equal
\u2234 Each angle will be = 90\u22182 = 45\u00b0 (b)<\/p>\n\n\n\n
An exterior angle of a triangle is equal to 100\u00b0 and two interior opposite angles are equal. Each of these angles is equal to
(a) 75\u00b0
(b) 80\u00b0
(c) 40\u00b0
(d) 50\u00b0
Solution:<\/strong>
In a triangle, exterior angles is equal to the sum of its interior opposite angles
\u2234 Sum of interior opposite angles = 100\u00b0
\u2235 Both angles are equal
\u2234 Each angle will be = 100\u22182 = 50\u00b0 (d)<\/p>\n\n\n\n
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:<\/strong>
Let \u2220A, \u2220B, \u2220C be the angles of a \u2206ABC and let \u2220A = \u2220B + \u2220C
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771526488_2d8ddd2a38_o.png\")
<\/figure>\n\n\n\n
But \u2220A + \u2220B + \u2220C = 180\u00b0
( Sum of angles of a triangle)
\u2234 \u2220A + \u2220A = 180\u00b0 \u21d2 2\u2220A = 180\u00b0
\u21d2 \u2220A = 180\u22182 = 90\u00b0
\u2234 \u2206 is a right triangle (d)<\/p>\n\n\n\n
Side BC of a triangle ABC has been produced to a point D such that \u2220ACD = 120\u00b0. If \u2220B = 12\u2220A, then \u2220A is equal to
(a) 80\u00b0
(b) 75\u00b0
(c) 60\u00b0
(d) 90\u00b0
Solution:<\/strong>
Side BC of \u2206ABC is produced to D, then
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771526298_f7872ee4f0_o.png\")
<\/figure>\n\n\n\n
Ext. \u2220ACB = \u2220A + \u2220B
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771526078_a4c500df3b_o.png\")
<\/figure>\n\n\n\n
In \u2206ABC, \u2220B = \u2220C and ray AX bisects the exterior angle \u2220DAC. If \u2220DAX = 70\u00b0, then \u2220ACB =
(a) 35\u00b0
(b) 90\u00b0
(c) 70\u00b0
(d) 55\u00b0
Solution:<\/strong>
In \u2206ABC, \u2220B = \u2220C
AX is the bisector of ext. \u2220CAD
\u2220DAX = 70\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771525848_a55c843599_o.png\")
<\/figure>\n\n\n\n
\u2234 \u2220DAC = 70\u00b0 x 2 = 140\u00b0
But Ext. \u2220DAC = \u2220B + \u2220C
= \u2220C + \u2220C (\u2235 \u2220B = \u2220C)
= 2\u2220C
\u2234 2\u2220C = 140\u00b0 \u21d2 \u2220C = 140\u22182 = 70\u00b0
\u2234 \u2220ACB = 70\u00b0 (c)<\/p>\n\n\n\n
In a triangle, an exterior angle at a vertex is 95\u00b0 and its one of the interior opposite angle is 55\u00b0, then the measure of the other interior angle is
(a) 55\u00b0
(b) 85\u00b0
(c) 40\u00b0
(d) 9.0\u00b0
Solution:<\/strong>
In \u2206ABC, BA is produced to D such that \u2220CAD = 95\u00b0
and let \u2220C = 55\u00b0 and \u2220B = x\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771525548_0d22741b02_o.png\")
<\/figure>\n\n\n\n
\u2235 Exterior angle of a triangle is equal to the sum of its opposite interior angle
\u2234 \u2220CAD = \u2220B + \u2220C \u21d2 95\u00b0 = x + 55\u00b0
\u21d2 x = 95\u00b0 \u2013 55\u00b0 = 40\u00b0
\u2234 Other interior angle = 40\u00b0 (c)<\/p>\n\n\n\n
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
(a) 90\u00b0
(b) 180\u00b0
(c) 270\u00b0
(d) 360\u00b0
Solution:<\/strong>
In \u2206ABC, sides AB, BC and CA are produced in order, then exterior \u2220FAB, \u2220DBC and \u2220ACE are formed
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771525318_3697f42151_o.png\")
<\/figure>\n\n\n\n
We know an exterior angles of a triangle is equal to the sum of its interior opposite angles
\u2234 \u2220FAB = \u2220B + \u2220C
\u2220DBC = \u2220C + \u2220A and
\u2220ACE = \u2220A + \u2220B Adding we get,
\u2220FAB + \u2220DBC + \u2220ACE = \u2220B + \u2220C + \u2220C + \u2220A + \u2220A + \u2220B
= 2(\u2220A + \u2220B + \u2220C)
= 2 x 180\u00b0 (Sum of angles of a triangle)
= 360\u00b0 (d)<\/p>\n\n\n\n
In \u2206ABC, if \u2220A = 100\u00b0, AD bisects \u2220A and AD\u22a5 BC. Then, \u2220B =
(a) 50\u00b0
(b) 90\u00b0
(c) 40\u00b0
(d) 100\u00b0
Solution:<\/strong>
In \u2206ABC, \u2220A = 100\u00b0
AD is bisector of \u2220A and AD \u22a5 BC
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771525078_187e994d2b_o.png\")
<\/figure>\n\n\n\n
Now, \u2220BAD = 100\u22182 = 50\u00b0
In \u2206ABD,
\u2220BAD + \u2220B + \u2220D= 180\u00b0
(Sum of angles of a triangle)
\u21d2 \u222050\u00b0 + \u2220B + 90\u00b0 = 180\u00b0
\u2220B + 140\u00b0 = 180\u00b0
\u21d2 \u2220B = 180\u00b0 \u2013 140\u00b0 \u2220B = 40\u00b0 (c)<\/p>\n\n\n\n
An exterior angle of a triangle is 108\u00b0 and its interior opposite angles are in the ratio 4:5. The angles of the triangle are
(a) 48\u00b0, 60\u00b0, 72\u00b0
(b) 50\u00b0, 60\u00b0, 70\u00b0
(c) 52\u00b0, 56\u00b0, 72\u00b0
(d) 42\u00b0, 60\u00b0, 76\u00b0
Solution:<\/strong>
In \u2206ABC, BC is produced to D and \u2220ACD = 108\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771524628_b32f57a76b_o.png\")
<\/figure>\n\n\n\n
Ratio in \u2220A : \u2220B = 4:5
\u2235 Exterior angle of a triangle is equal to the sum of its opposite interior angles
\u2234 \u2220ACD = \u2220A + \u2220B = 108\u00b0
Ratio in \u2220A : \u2220B = 4:5
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771524898_06d6d23cb6_o.png\")
<\/figure>\n\n\n\n
In a \u2206ABC, if \u2220A = 60\u00b0, \u2220B = 80\u00b0 and the bisectors of \u2220B and \u2220C meet at O, then \u2220BOC =
(a) 60\u00b0
(b) 120\u00b0
(c) 150\u00b0
(d) 30\u00b0
Solution:<\/strong>
In \u2206ABC, \u2220A = 60\u00b0, \u2220B = 80\u00b0
\u2234 \u2220C = 180\u00b0 \u2013 (\u2220A + \u2220B)
= 180\u00b0 \u2013 (60\u00b0 + 80\u00b0)
= 180\u00b0 \u2013 140\u00b0 = 40\u00b0
Bisectors of \u2220B and \u2220C meet at O
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771524508_00675cf207_o.png\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771524338_04cca3e841_o.png\")
<\/figure>\n\n\n\n
Line segments AB and CD intersect at O such that AC || DB. If \u2220CAB = 45\u00b0 and \u2220CDB = 55\u00b0, then \u2220BOD =
(a) 100\u00b0
(b) 80\u00b0
(c) 90\u00b0
(d) 135\u00b0
Solution:<\/strong>
In the figure,
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771524098_fb2d48cb8d_o.png\")
<\/figure>\n\n\n\n
AB and CD intersect at O
and AC || DB, \u2220CAB = 45\u00b0
and \u2220CDB = 55\u00b0
\u2235 AC || DB
\u2234 \u2220CAB = \u2220ABD (Alternate angles)
In \u2206OBD,
\u2220BOD = 180\u00b0 \u2013 (\u2220CDB + \u2220ABD)
= 180\u00b0 \u2013 (55\u00b0 + 45\u00b0)
= 180\u00b0 \u2013 100\u00b0 = 80\u00b0 (b)<\/p>\n\n\n\n
In the figure, if EC || AB, \u2220ECD = 70\u00b0 and \u2220BDO = 20\u00b0, then \u2220OBD is
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771523918_bb7d8369e3_o.png\")
<\/figure>\n\n\n\n
(a) 20\u00b0
(b) 50\u00b0
(c) 60\u00b0
(d) 70\u00b0
Solution:<\/strong>
In the figure, EC || AB
\u2220ECD = 70\u00b0, \u2220BDO = 20\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771523768_fcea2cc549_o.png\")
<\/figure>\n\n\n\n
\u2235 EC || AB
\u2220AOD = \u2220ECD (Corresponding angles)
\u21d2 \u2220AOD = 70\u00b0
In \u2206OBD,
Ext. \u2220AOD = \u2220OBD + \u2220BDO
70\u00b0 = \u2220OBD + 20\u00b0
\u21d2 \u2220OBD = 70\u00b0 \u2013 20\u00b0 = 50\u00b0 (b)<\/p>\n\n\n\n
In the figure, x + y =
(a) 270
(b) 230
(c) 210
(d) 190\u00b0
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771523578_0f93a80d33_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure
<\/p>\n\n\n\n<\/figure>\n\n\n\n
Ext. \u2220OAE = \u2220AOC + \u2220ACO
\u21d2 x = 40\u00b0 + 80\u00b0 = 120\u00b0
Similarly,
Ext. \u2220DBF = \u2220ODB + \u2220DOB
y = 70\u00b0 + \u2220DOB
[(\u2235 \u2220AOC = \u2220DOB) (vertically opp. angles)]
= 70\u00b0 + 40\u00b0 = 110\u00b0
\u2234 x+y= 120\u00b0+ 110\u00b0 = 230\u00b0 (b)<\/p>\n\n\n\n
If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?
(a) 25\u00b0
(b) 30\u00b0
(c) 45\u00b0
(d) 60\u00b0
Solution:<\/strong>
Ratio in the measures of the triangle =3:4:5
Sum of angles of a triangle = 180\u00b0
Let angles be 3x, 4x, 5x
Sum of angles = 3x + 4x + 5x = 12x
\u2234 Smallest angle = 180x3x12x = 45\u00b0 (c)<\/p>\n\n\n\n
In the figure, if AB \u22a5 BC, then x =
(a) 18
(b) 22
(c) 25
(d) 32
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771523138_c86807b0a9_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, AB \u22a5 BC
\u2220AGF = 32\u00b0
\u2234 \u2220CGB = \u2220AGF (Vertically opposite angles)
= 32\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771522888_b62f5c3e0b_o.png\")
<\/figure>\n\n\n\n
In \u2206GCB, \u2220B = 90\u00b0
\u2234 \u2220CGB + \u2220GCB = 90\u00b0
\u21d2 32\u00b0 + \u2220GCB = 90\u00b0
\u21d2 \u2220GCB = 90\u00b0 \u2013 32\u00b0 = 58\u00b0
Now in \u2206GDC,
Ext. \u2220GCB = \u2220CDG + \u2220DGC
\u21d2 58\u00b0 = x + 14\u00b0 + x
\u21d2 2x + 14\u00b0 = 58\u00b0
\u21d2 2x = 58 \u2013 14\u00b0 = 44
\u21d2 x = 442 = 22\u00b0
\u2234 x = 22\u00b0 (b)<\/p>\n\n\n\n
In the figure, what is \u2220 in terms of x and y?
(a) x + y + 180
(b) x + y \u2013 180
(c) 180\u00b0 -(x+y)
(d) x+y + 360\u00b0
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771522658_fc4cf24a86_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, BC is produced both sides CA and BA are also produced
In \u2206ABC,
\u2220B = 180\u00b0 -y
and \u2220C 180\u00b0 \u2013 x
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771522378_1181db9809_o.png\")
<\/figure>\n\n\n\n
\u2234 z = \u2220A = 180\u00b0 \u2013 (B + C)
= 180\u00b0 \u2013 (180 \u2013 y + 180 -x)
= 180\u00b0 \u2013 (360\u00b0 \u2013 x \u2013 y)
= 180\u00b0 \u2013 360\u00b0 + x + y = x + y \u2013 180\u00b0 (b)<\/p>\n\n\n\n
In the figure, for which value of x is l1<\/sub> || l2<\/sub>?
(a) 37
(b) 43
(c) 45
(d) 47
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771522158_03d94f2e14_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, l1<\/sub> || l2<\/sub>
\u2234 \u2220EBA = \u2220BAH (Alternate angles)
\u2234 \u2220BAH = 78\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771521998_5b82383db2_o.png\")
<\/figure>\n\n\n\n
\u21d2 \u2220BAC + \u2220CAH = 78\u00b0
\u21d2 \u2220BAC + 35\u00b0 = 78\u00b0
\u21d2 \u2220BAC = 78\u00b0 \u2013 35\u00b0 = 43\u00b0
In \u2206ABC, \u2220C = 90\u00b0
\u2234 \u2220ABC + \u2220BAC = 90\u00b0
\u21d2 x + 43\u00b0 = 90\u00b0 \u21d2 x = 90\u00b0 \u2013 43\u00b0
\u2234 x = 47\u00b0 (d)<\/p>\n\n\n\n
In the figure, what is y in terms of x?
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771521978_bfcc734741_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In \u2206ABC,
\u2220ACB = 180\u00b0 \u2013 (x + 2x)
= 180\u00b0 \u2013 3x \u2026(i)
and in \u2206BDG,
\u2220BED = 180\u00b0 \u2013 (2x + y) \u2026(ii)
\u2220EGC = \u2220AGD (Vertically opposite angles)
= 3y
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771521488_40f9e97c51_o.png\")
<\/figure>\n\n\n\n
In quad. BCGE,
\u2220B + \u2220ACB + \u2220CGE + \u2220BED = 360\u00b0 (Sum of angles of a quadrilateral)
\u21d2 2x+ 180\u00b0 \u2013 3x + 3y + 180\u00b0- 2x-y = 360\u00b0
\u21d2 -3x + 2y = 0
\u21d2 3x = 2y \u21d2 y = 32x (a)<\/p>\n\n\n\n
In the figure, what is the value of x?
(a) 35
(b) 45
(c) 50
(d) 60
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771521268_5cd454812e_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, side AB is produced to D
\u2234 \u2220CBA + \u2220CBD = 180\u00b0 (Linear pair)
\u21d2 7y + 5y = 180\u00b0
\u21d2 12y = 180\u00b0
\u21d2 y = 18012 = 15
and Ext. \u2220CBD = \u2220A + \u2220C
\u21d2 7y = 3y + x
\u21d2 7y -3y = x
\u21d2 4y = x
\u2234 x = 4 x 15 = 60 (d)<\/p>\n\n\n\n
In the figure, the value of x is
(a) 65\u00b0
(b) 80\u00b0
(c) 95\u00b0
(d) 120\u00b0
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771521088_9e33e185b6_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, \u2220A = 55\u00b0, \u2220D = 25\u00b0 and \u2220C = 40\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771520978_ddeab49558_o.png\")
<\/figure>\n\n\n\n
Now in \u2206ABD,
Ext. \u2220DBC = \u2220A + \u2220D
= 55\u00b0 + 25\u00b0 = 80\u00b0
Similarly, in \u2206BCE,
Ext. \u2220DEC = \u2220EBC + \u2220ECB
= 80\u00b0 + 40\u00b0 = 120\u00b0 (d)<\/p>\n\n\n\n
In the figure, if BP || CQ and AC = BC, then the measure of x is
(a) 20\u00b0
(b) 25\u00b0
(c) 30\u00b0
(d) 35\u00b0
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771520848_d8afc5e31c_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure, AC = BC, BP || CQ
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771520668_50044bd710_o.png\")
<\/figure>\n\n\n\n
\u2235 BP || CQ
\u2234 \u2220PBC \u2013 \u2220QCD
\u21d2 20\u00b0 + \u2220ABC = 70\u00b0
\u21d2 \u2220ABC = 70\u00b0 \u2013 20\u00b0 = 50\u00b0
\u2235 BC = AC
\u2234 \u2220ACB = \u2220ABC (Angles opposite to equal sides)
= 50\u00b0
Now in \u2206ABC,
Ext. \u2220ACD = \u2220B + \u2220A
\u21d2 x + 70\u00b0 = 50\u00b0 + 50\u00b0
\u21d2 x + 70\u00b0 = 100\u00b0
\u2234 x = 100\u00b0 \u2013 70\u00b0 = 30\u00b0 (c)<\/p>\n\n\n\n
In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If \u2220APR = 25\u00b0, \u2220RQC = 30\u00b0 and \u2220CQF = 65\u00b0, then
(a) x = 55\u00b0, y = 40\u00b0
(b) x = 50\u00b0, y = 45\u00b0
(c) x = 60\u00b0, y = 35\u00b0
(d) x = 35\u00b0, y = 60\u00b0
<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771520458_6a960b38ba_o.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In the figure,
\u2235 AB || CD, EF intersects them at P and Q respectively,
\u2220APR = 25\u00b0, \u2220RQC = 30\u00b0, \u2220CQF = 65\u00b0
\u2235 AB || CD
\u2234 \u2220APQ = \u2220CQF (Corresponding anlges)
\u21d2 y + 25\u00b0 = 65\u00b0
\u21d2 y = 65\u00b0 \u2013 25\u00b0 = 40\u00b0
and APQ + PQC = 180\u00b0 (Co-interior angles)
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771520308_2cb60e2410_o.png\")
<\/figure>\n\n\n\n
y + 25\u00b0 + \u22201 +30\u00b0= 180\u00b0
40\u00b0 + 25\u00b0 + \u22201 + 30\u00b0 = 180\u00b0
\u21d2 \u22201 + 95\u00b0 = 180\u00b0
\u2234 \u22201 = 180\u00b0 \u2013 95\u00b0 = 85\u00b0
Now, \u2206PQR,
\u2220RPQ + \u2220PQR + \u2220PRQ = 180\u00b0 (Sum of angles of a triangle)
\u21d2 40\u00b0 + x + 85\u00b0 = 180\u00b0
\u21d2 125\u00b0 + x = 180\u00b0
\u21d2 x = 180\u00b0 \u2013 125\u00b0 = 55\u00b0
\u2234 x = 55\u00b0, y = 40\u00b0 (a)<\/p>\n\n\n\n
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94\u00b0 and 126\u00b0. Then, \u2220BAC = ?
(a) 94\u00b0
(b) 54\u00b0
(c) 40\u00b0
(d) 44\u00b0
Solution:<\/strong>
In \u2206ABC, base BC is produced both ways and \u2220ACD = 94\u00b0, \u2220ABE = 126\u00b0
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/31771520108_b305ef5a4e_o.png\")
<\/figure>\n\n\n\n
Ext. \u2220ACD = \u2220BAC + \u2220ABC
\u21d2 94\u00b0 = \u2220BAC + \u2220ABC
Similarly, \u2220ABE = \u2220BAC + \u2220ACB
\u21d2 126\u00b0 = \u2220BAC + \u2220ACB
Adding,
94\u00b0 + 126\u00b0 = \u2220BAC + \u2220ABC + \u2220ACB + \u2220BAC
220\u00b0 = 180\u00b0 + \u2220BAC
\u2234 \u2220BAC = 220\u00b0 -180\u00b0 = 40\u00b0 (c)<\/p>\n\n\n\n