Sol :
Two<\/p>\n\n\n\n
(ii) In figure (i) , if \u2220ACD=130\u00b0 , \u2220ABC=48\u00b0 , then \u2220BAC=__<\/strong> (iii) If a side of a triangle is extended , then the exterior angle so formed is equal to the __ of the opposite interior angles .<\/strong> Which of the following statements are true (T) and which are false (F):<\/strong> (ii) Sum of the four angles of a quadrilateral is three right angles.<\/strong>F<\/p>\n\n\n\n (iii) In a triangle, there may be two right angles.<\/strong>F (i) In a triangle, one angle measures 70\u00b0,then write the sum of two remaining angles in degrees<\/strong> (iii) In a \u0394ABC , \u2220B=105\u00b0 and \u2220C=50\u00b0 , find the value of \u2220A <\/strong> (ii) \u2220A+\u2220B+\u2220C=180\u00b0(sum of angles of triangle is 180\u00b0) (iii) \u2220A+\u2220B+\u2220C=180\u00b0 Give reasons for your answer in each of the following cases. Can in a triangle there be ?<\/strong> (i) At least how many acute angles are there in a triangle ?<\/strong> (iii) At most how many obtuse angles can there be in a triangle ?<\/strong> (i) If \u03b1 be an angle of a triangle such that 90\u00b0<\u03b1<180\u00b0 , then what type of triangle will it be ?<\/strong> (i) In a right angled triangle, one acute angle is 40\u00b0 , then find the measure of the other acute angle in degrees.<\/strong> How many right angles is the sum of the interior angles of a rhombus ?<\/strong> (i) In the given f\u200cigure side BC is produced to M, \u2220ACM=100\u00b0 and \u2220ABC=45\u00b0 , find \u2220BAC.<\/strong> (ii) In the given f\u200cigure, if \u2220MBC=140\u00b0 , \u2220BCA=40\u00b0 , find \u2220BAC.<\/strong> Sol : (iii) If \u03b8 be the exterior angle of a triangle and sum of the two opposite interior angles, one is \u03b2 , then what is the relation between \u03b8 and \u03b2 ?<\/strong> (iv) If in \u0394ABC (figure below) , \u2220ACD=105\u00b0 and \u2220BAC=35\u00b0 , find the value of \u2220ABC.<\/strong> (i)<\/strong> In the following f\u200cigures, f\u200cind the values of each angle. \u2220A=4x \u2220B=2x \u2220C=3x (b)<\/strong> Sol:<\/p>\n\n\n\n \u2220B=4x<\/p>\n\n\n\n =4\u00d736\u00b0<\/p>\n\n\n\n =144\u00b0<\/p>\n\n\n\n (ii) In the given f\u200cigure, AM and DM are the bisectors of \u2220A and \u2220D respectively of quadrilateral ABCD . Find the value of \u2220AMD in degrees.<\/strong> \u2220MAD=\\frac{1}{2}\u2220A In quadrilateral ABCD<\/p>\n\n\n\n \u2220A+\u2220B+\u2220C+\u2220D=360\u00b0<\/p>\n\n\n\n 2\u2220MAD+50\u00b0+100\u00b0+2\u2220MDA=360\u00b0<\/p>\n\n\n\n 2(\u2220MAD+\u2220MDA)+150\u00b0=360\u00b0<\/p>\n\n\n\n 2(\u2220MAD+\u2220MDA)=360\u00b0-150\u00b0 \u0394MDA Page 8.45<\/p>\n\n\n\n (i) In a triangle, two angles are equal and third angle exceeds each of these angles by 30\u00b0. Determine all angles of the triangle<\/strong> \u2220A=x+30\u00b0 \u2220B=x=50\u00b0<\/p>\n\n\n\n \u2220C=x=50\u00b0<\/p>\n\n\n\n (ii)<\/strong> Find the angle formed by the bisectors of two acute of a right-angled triangle.<\/strong> (i) If the angles of a triangle are in the ratio 2:3:4 , then find the values of the biggest and smallest angles.<\/strong> Type 2<\/p>\n\n\n\n (i) In figure (i) below , side QR of a \u0394PQR is produced to S . If \u2220P:\u2220Q:\u2220R=3:2:1 and RT\u22a5PR , find \u2220TRS<\/strong> (ii) In quadrilateral ABCD, AD||BC and bisectors of interior \u2220A and \u2220B meet at O , find the measure of \u2220AOB in degree [fig. (ii)]<\/strong> (iii) In figure (iii) , AB||CD , then find the value of \u03b1+\u03b2+\u03b3.<\/strong> (iv) In figure (iv) , l||m , find the value of x<\/strong> (vi) In figure (a) and (b) , l||m , then find the value of x <\/strong> (b) (vii) In f\u200cigure (a) and (b) below, f\u200cind x and y as required.<\/strong> (b) (viii) In the given f\u200cigure, sides BC, CA and BA of \u0394ABC are produced to D, Q and P respectively. If \u2220ACD=100\u00b0 and \u2220QAP=35\u00b0 , then f\u200cind all angles of the triangle. (ix) In \u0394ABC , \u2220B=45\u00b0 , \u2220C=55\u00b0 and bisector of \u2220A meets BC in D. Find \u2220ADB and \u2220ADC Type 3<\/p>\n\n\n\n In the given figure , prove that P||m<\/strong> Prove that the sum of the exterior angles formed on producing sides of a triangle in the same order is 360\u00b0 or four right angles .<\/strong> Prove that each angle of a triangle will be 60\u00b0, if all its angles are of the same measure.<\/strong> If an angle of a triangle is equal to the sum of remaining two angles, prove that triangle is right angled triangle.<\/strong> In \u0394ABC,\u2220A = 40\u00b0, if bisectors of \u2220B and \u2220C meet at point O, then prove \u2220BOC=110\u00b0<\/strong> In right angled \u0394ABC, \u2220A=90\u00b0 , and bisectors of \u2220B and \u2220C meet at O, prove that \u2220BOC = 135\u00b0<\/strong> In a triangle ABC, \u2220A=90\u00b0 and AL\u22a5BC , prove that \u2220BAL=\u2220ACB <\/strong> If a transversal intersects two parallel lines, prove that bisectors of two interior angles on the same side form 90\u00b0 with one another.<\/strong> Prove that the sum of interior angles of a hexagon is 720\u00b0<\/strong> In a triangle ABC , CD is perpendicular to side AB and BE is perpendicular to side AC. Prove that \u2220ABE=\u2220ACD<\/strong><\/p>\n\n\n\n [Hint: In \u0394ABE and \u0394ADC In \u0394ABC , internal bisectors of \u2220B and \u2220C meet at P and external bisectors of these angles meet at Q. Prove that<\/strong> In the given figure (i) , two plane mirrors m<\/em> and n<\/em> are perpendicular to each other. Show that incident ray CA on being reflected is parallel to reflected ray BD.<\/strong> In the given f\u200cigure, PS is bisector of \u2220P and PT\u22a5QR , prove that <\/strong> In the given f\u200cigure, side BC of \u0394ABC is produced so as to form the ray BD and CE||AB. Without using the theorem related to sum of angles of a triangle, prove that:<\/strong> In \u0394ABC , side BC is produced to D and bisector of \u2220A meets BC at L. Prove that \u2220ABC+\u2220ACD=2\u2220ALC<\/strong> ABCDE is a regular pentagon and bisector of \u2220A meets the side CD at point . Prove that \u2220AMC=90\u00b0<\/strong> In the given f\u200cigure, AE bisects \u2220CAD and \u2220B=\u2220C . Prove that AE||BC<\/strong> If arms of an angle are perpendicular to the arms of another angle, then prove that the angles will either be equal or supplementary<\/strong> In the given f\u200cigure, bisectors of \u2220B and \u2220D meet produced CD and AB at P and Q respectively.<\/strong>
Sol :
82\u00b0<\/p>\n\n\n\n
Sol :
Sum<\/p>\n\n\n\n
\n\n\n\nQuestion 2<\/h4>\n\n\n\n
(i) Sum of the three angles of a triangles is 180\u00b0.<\/strong>T<\/p>\n\n\n\n
(iv) In a triangle, exterior angle is smaller than each interior opposite angle.<\/strong>F
(v) In a triangle, there may be two obtuse angles.<\/strong>F
(vi) In a triangle, there may be two acute angles.<\/strong>T
(vii) In a triangle, exterior angle is equal to the sum of interior opposite angles.<\/strong>T<\/p>\n\n\n\n
\n\n\n\nQuestion 3<\/h4>\n\n\n\n
(ii) In a \u0394ABC , \u2220C=40\u00b0 ,\u2220B=80\u00b0 , find the value of \u2220A <\/strong><\/p>\n\n\n\n
Sol :
(i) The sum of two remaining angle
=180\u00b0-70\u00b0
=110\u00b0<\/p>\n\n\n\n
\u2220A+80\u00b0+40\u00b0=180\u00b0
\u2220A+120\u00b0=180\u00b0
\u2220A=60\u00b0<\/p>\n\n\n\n
\u2220A+105\u00b0+50\u00b0=180\u00b0
\u2220A+155\u00b0=180\u00b0
\u2220A=25\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 4<\/h4>\n\n\n\n
(i) Two right angles<\/strong>
(ii) Two obtuse angles<\/strong>
(iii) Two acute angles<\/strong>
(iv) Each angle greater than 60\u00b0<\/strong>
(v) Each angle smaller than 60\u00b0<\/strong>
(vi) Each angle equal to 60\u00b0<\/strong><\/p>\n\n\n\n
\n\n\n\nQuestion 5<\/h4>\n\n\n\n
(ii) What Will be the maximum numbers of acute angles in a triangles ?<\/strong><\/p>\n\n\n\n
Sol :<\/p>\n\n\n\n![\"\"\/](\"https:\/\/scontent.fdel3-2.fna.fbcdn.net\/v\/t1.15752-9\/p1080x2048\/99060933_563671571247178_8328501594844823552_n.jpg?_nc_cat=110&_nc_sid=b96e70&_nc_ohc=zSNVw0EhVMoAX_XWsqv&_nc_ht=scontent.fdel3-2.fna&_nc_tp=6&oh=9077629389f3e7cdc5ca1ecd997d88a9&oe=5F091363\")
<\/a><\/figure>\n\n\n\n
\n\n\n\nQuestion 6<\/h4>\n\n\n\n
(ii) In a right angled triangle, how many acute angles are there ?<\/strong><\/p>\n\n\n\n
\n\n\n\nQuestion 7<\/h4>\n\n\n\n
(ii) A quadrilateral has three angles as 110\u00b0 , 40\u00b0 and 50\u00b0 . Find the value of the fourth angle.<\/strong><\/p>\n\n\n\n
\n\n\n\nQuestion 8<\/h4>\n\n\n\n
Sol :
Four right angles<\/p>\n\n\n\n
\n\n\n\nQuestion 9<\/h4>\n\n\n\n
Sol :
\u2220BAC+\u2220ABC=\u2220ACM
\u2220BAC+45\u00b0=100\u00b0
\u2220BAC=55\u00b0<\/p>\n\n\n\n
We know that exterior angle is equal to the sum of opposite interior angles.
\u2220MBC=\u2220BAC+\u2220BCA
140\u00b0=\u2220BAC+40\u00b0
\u2220BAC=140\u00b0-40\u00b0
\u2220BAC=100\u00b0<\/p>\n\n\n\n
Sol :
\u03b8>\u03b2<\/p>\n\n\n\n
<fig to be added>
Sol :
We know that in triangle exterior angle is equal to the sum of opposite interior angles.
\u2220ACD=\u2220BAC+\u2220ABC
105\u00b0=35\u00b0+\u2220ABC
\u2220ABC=105\u00b0-35\u00b0
\u2220ABC=70\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 10<\/h4>\n\n\n\n
(a)<\/strong>
Sol :
\u2220A+\u2220B+\u2220C=180\u00b0
4x+2x+3x=180\u00b0
9x=180\u00b0
x=\\frac{180}{9}
x=20\u00b0<\/p>\n\n\n\n
=4\u00d720\u00b0
=80\u00b0<\/p>\n\n\n\n
=2\u00d720\u00b0
=40\u00b0<\/p>\n\n\n\n
=3\u00d720\u00b0
=60\u00b0<\/p>\n\n\n\n
<\/a><\/p>\n\n\n\n
<fig to be added><\/strong>
Sol :
AM and DM are the bisector of \u2220A and \u2220D<\/p>\n\n\n\n
\u21d22\u2220MAD=\u2220A
\\angle MDA=\\frac{1}{2} \\angle D
\u21d22\u2220MDA=\u2220D<\/p>\n\n\n\n
\u2220MAD+\u2220MDA=\\frac{210^{\\circ}}{2}
\u2220MAD+\u2220MDA=105\u00b0<\/p>\n\n\n\n
\u2220MAD+\u2220MDA+\u2220AMD=180\u00b0
105\u00b0+\u2220AMD=180\u00b0
\u2220AMD=75\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 11<\/h4>\n\n\n\n
(ii) In a triangle, an exterior angle is 115\u00b0 and one of the opposite interior angles 35\u00b0 , then find the remaining angles <\/strong>
(iii) An angle of a triangle measures 65\u00b0 , then f\u200cind the other two angles if their difference is 25\u00b0<\/strong>
(iv) In a right angled triangle, the bigger acute angle is two times the smaller acute angle, then f\u200cind the measure of bigger acute angle.<\/strong>
(v) In a triangle, one of the three angles twice the smallest angle and other is three times the smallest angle. Find the angles.<\/strong>
(vi) Sum of two angles of a triangle is 80\u00b0 and their difference is 20\u00b0 . Find all angles of the triangle .<\/strong>
Sol :
(i)<\/strong>
Diagram
\u2220B=x ,\u2220C=x,\u2220A=x+30\u00b0
\u2220A+\u2220B+\u2220C=180\u00b0
x+30+x+x=180\u00b0
3x+30\u00b0=180\u00b0
3x=150\u00b0
x=\\frac{150}{3}
x=50\u00b0<\/p>\n\n\n\n
=50\u00b0+30\u00b0
=80\u00b0<\/p>\n\n\n\n
Diagram
\u2220BAC+\u2220ABC=\u2220ACD
\u2220BAC+35\u00b0=115\u00b0
\u2220BAC=80\u00b0
\u2220ACB+\u2220ACD=180\u00b0(linear pair)
\u2220ACB+115\u00b0=180\u00b0
\u2220ACB=65\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 12<\/h4>\n\n\n\n
Sol :<\/p>\n\n\n\n
\n\n\n\nQuestion 13<\/h4>\n\n\n\n
(ii) If the angles of a triangle are in the ratio 1:2:3 , then find the value of the greater angle of the triangles.<\/strong>
(iii) If angles of a triangle are in the ratio 2:3:5 , then f\u200cind the measure in degree of the smallest angle.<\/strong>
(iv) if angles of a quadrilateral are in the ratio 1:2:3:4 , then find the angles of the quadrilateral<\/strong><\/p>\n\n\n\n
\n\n\n\nQuestion 14<\/h4>\n\n\n\n
Sol :<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
Sol :<\/p>\n\n\n\n
[Hint : Draw a line OE through point O parallel to AB or CD , then \u2220BOE=180\u00b0-\u03b1 [alternate angles]
\u2220DOE=180\u00b0-\u03b2 [alternate angle]
Now , \u2220BOE+\u2220DOE=\u03b3=180\u00b0-\u03b1+180\u00b0-\u03b2
or \u03b1+\u03b2+\u03b3=360\u00b0]<\/p>\n\n\n\n
<fig to be added>
[Hint: l||m and they are intersected by a transversal
\u2234 60\u00b0+y=180\u00b0\u21d2y=180\u00b0-60\u00b0=120\u00b0
\u2234 x=y+50\u00b0=120\u00b0+50\u00b0=170\u00b0]
(v) In figure (v) , ABC is an isosceles triangles whose side AB=AC and XY||BC. If \u2220A=30\u00b0 , then find the value if \u2220BXY in degrees.<\/strong>
[Hint: AB=AC [\u2234 \u0394ABC is an isoceles triangle]
\u2234 \u2220BXY+\u2220ABC=180\u00b0 or \u2220BXY + 75\u00b0 = 180\u00b0
\u2234 \u2220BXY=180\u00b0-75\u00b0 = 105\u00b0 ]<\/p>\n\n\n\n
(a)
(a)
[Hint: In figure (a)
From \u0394AEC, x=30\u00b0+42\u00b0=72\u00b0
Also, y=180\u00b0-42\u00b0=138\u00b0
In figure (b) exterior \u2220CBD=30\u00b0+34\u00b0=64\u00b0=\u2220EBD
\u2234 Exterior angle x=\u2220EBD+\u2220EDB=64\u00b0+45\u00b0=109\u00b0 ]<\/p>\n\n\n\n
[Hint: \u2220BAC=\u2220QAP=35\u00b0
Now , exterior \u2220DCA=\u2220CAB+\u2220ABC
\u21d2100\u00b0=35\u00b0+\u2220ABC or \u2220ABC=100\u00b0-35\u00b0=65\u00b0
Also , \u2220ACB=180\u00b0-100\u00b0=80\u00b0 ]<\/p>\n\n\n\n
[Hint: \u2220A=180\u00b0-(45\u00b0+55\u00b0)=80\u00b0
\u2235 AD is the bisector of \u2220A ,
\u2234 \u2220BAD=\\angle DAC=\\dfrac{80^{\\circ}}{2}=40^{\\circ}
Now exterior \u2220ADB=\u2220DAC+\u2220DCA=40\u00b0+55\u00b0=95\u00b0
and exterior \u2220ADC=\u2220ABD+\u2220BAD=45\u00b0+40\u00b0=85\u00b0 ]<\/p>\n\n\n\n
\n\n\n\nQuestion 15<\/h4>\n\n\n\n
\n\n\n\nQuestion 16<\/h4>\n\n\n\n
Sol :<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
\n\n\n\nQuestion 17<\/h4>\n\n\n\n
Sol :<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n<\/a><\/figure>\n\n\n\n
\n\n\n\nQuestion 18<\/h4>\n\n\n\n
Sol :<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n<\/a><\/figure>\n\n\n\n
\n\n\n\nQuestion 19<\/h4>\n\n\n\n
Sol :
[Hint: See the given f\u200cigure
Since BO and CO are bisectors of \u2220ABC
and \u2220ACB respectively ,
\u2234 \u2220ABO=\u2220OBC=\u22201 (say)
and \u2220ACO=\u2220OCB=\u22202 (say)
Now , in \u0394OBC,
\u22201+\u22202+\u2220BOC=180\u00b0
In \u0394ABC , \u2220ABC+\u2220ACB+\u2220BAC=180\u00b0..(i)
or 2\u22201+2\u22202+40\u00b0=180\u00b0..(ii) [\u2235 \u2220BAC=40\u00b0]
\u2234 \u22201+\u22202=\\dfrac{180^{\\circ}-40^{\\circ}}{2}=70^{\\circ}
\u2234 From (1) , 70\u00b0+\u2220BOC=180\u00b0
\u2220BOC=180\u00b0-70\u00b0=110\u00b0]<\/p>\n\n\n\n
\n\n\n\nQuestion 20<\/h4>\n\n\n\n
[Hint: Proceed as in question 19
\u22201+\u22202+\u2220BOC=180\u00b0
Also , \u2220A+\u2220B+\u2220C=180\u00b0
90\u00b0+2\u22201+2\u22202=180\u00b0
\u21d2\u22201+\u22202=\\dfrac{180^{\\circ}-90^{\\circ}}{2}=45^{\\circ}
From (i) , \u2220BOC=180\u00b0-(\u22201+\u22202)
=180\u00b0-45\u00b0
=135\u00b0]<\/p>\n\n\n\n
\n\n\n\nQuestion 21<\/h4>\n\n\n\n
<fig to be added>
[Hint : Draw f\u200cigure yourself, in \u0394ABC , \u2220ABC ,\u2220CAB+\u2220B+\u2220ACB=180\u00b0..(i)
Now, in \u0394ALB , \u2220ALB=90\u00b0
\u2234 \u2220ALB+\u2220B+\u2220BAL=180\u00b0..(ii)
From (i) and (ii) , we get
\u2220ALB+\u2220B+\u2220BAL=\u2220CAB+\u2220B+\u2220ACB
\u21d290\u00b0+\u2220B+\u2220BAL=90\u00b0+\u2220B+\u2220ACB
\u21d2\u2220BAL=\u2220ACB ]<\/p>\n\n\n\n
\n\n\n\nQuestion 22<\/h4>\n\n\n\n
<fig to be added>
[Hint: AB||CD and they are intersected by transversal EF at M and N (see the given fig)
Since sum of the interior angles on the same side of transversal is 180\u00b0
\u2234 \u2220BMN+\u2220DNM=180\u00b0
\u21d2\\dfrac{1}{2}\\angle BMN+\\dfrac{1}{2}\\angle DNM=90^{\\circ}
Now MO and NO are bisectors of \u2220BMN and \u2220DNM respectively.
\u2234 \u2220OMN+\u2220ONM=90\u00b0 or \u22201-\u22202=90\u00b0
Now , \u22201+\u22202+\u2220MON=180\u00b0
\u21d290\u00b0+\u2220MON=180\u00b0
\u21d2\u2220MON=90\u00b0]<\/p>\n\n\n\n
\n\n\n\nQuestion 23<\/h4>\n\n\n\n
[Hint: We know that the sum of the angles subtended by sides at the centre of a polygon is 360\u00b0.
Angles subtended a side of a regular polygon at its centre =\\dfrac{360\u00b0}{n} where n is the number of sides which is here 6
\u2235 \u2220AOB =\\dfrac{360^{\\circ}}{6} = 60\u00b0; Also AO=BO=AB
\u2234 \u0394AOB is an equilateral triangle.
\u2234 \u0394AOB=\u2220OBA=\u2220OAB=60\u00b0
Similarly, \u0394OAF is also an equilateral triangle.
\u2234 \u2220OAB+\u2220OAF=60\u00b0+60\u00b0=120\u00b0=\u2220FAB=an interior angle
\u2234 Sum of all the interior angles of a hexagon =6\u00d7120\u00b0=720]<\/p>\n\n\n\n
\n\n\n\nQuestion 24<\/h4>\n\n\n\n
\u2220AEB=\u2220ADC; \u2220A=\u2220A
\u2234 \u2220ABE=\u2220ACD]<\/p>\n\n\n\n
\n\n\n\nQuestion 25<\/h4>\n\n\n\n
\u2220BPC+\u2220BQC=180\u00b0<\/strong>
[Hint: \u2220BPC=90\u00b0+\\dfrac{\\angle A}{2} and
\u2220BQC=90\u00b0-\\dfrac{\\angle A}{2}
\u2234 \u2220BPC+\u2220BQC
=90\u00b0+\\dfrac{\\angle A}{2}+90\u00b0-\\dfrac{\\angle A}{2}
=180\u00b0]<\/p>\n\n\n\n
\n\n\n\nQuestion 26<\/h4>\n\n\n\n
[Hint: AP and BP are normals to mirrors m<\/em> and n<\/em> respectively ]
Angle of incidence = Angle of ref\u200clection [Law of reflection]
Now in figure, (ii) \u2220CAP=\u2220PAB=x
\u2220ABP=\u2220PBD=y
In \u0394APB , \u2220APB=90\u00b0
\u2234 \u2220PAB+\u2220PBA=90\u00b0
\u2234 x+y=90\u00b0 or 2(x+y)=2\u00d790\u00b0
or 2x+2y=180\u00b0 or \u2220CAB+\u2220ABD=180\u00b0
Since these angles are on the same side of the transversal AB, therefore CA||BD<\/p>\n\n\n\n
\n\n\n\nQuestion 27<\/h4>\n\n\n\n
\\angle TPS=\\dfrac{1}{2}(\\angle Q-\\angle R)
[Hint: Let \u2220QPS=\u2220RPS=x and \u2220TPS=y
In right angled \u0394PQT , \u2220QPT+\u2220Q=(x-y)+\u2220Q=90\u00b0..(i)
In right angled \u0394PRT , \u2220RPT+\u2220R=90\u00b0 or
(x+y)+\u2220R=90\u00b0..(iii) [\u2235 \u2220RPT=\u2220RPS+\u2220TPS=x+y]
From (i) and (ii) , x-y+\u2220Q=x+y+\u2220R
\u21d2 \u2220Q-\u2220R=2y
\u2234 y=\u2220TPS=\\dfrac{1}{2}(\\angle Q – \\angle R)]<\/p>\n\n\n\n
\n\n\n\nQuestion 28<\/h4>\n\n\n\n
\u2220ACD=\u2220A+\u2220B and hence determine \u2220A+\u2220B+\u2220C=180\u00b0.<\/strong>
<fig to be added>
[Hint: \u2235 AB||CE and transversal AC meets them
\u2234 \u22201=\u22204 [alternate angles]
and \u22202=\u22205 [corresponding angles]
\u2234 \u22201+\u22202=\u22204+\u22205=\u2220ACE+\u2220ECD=\u2220ACD
\u2234 \u2220A+\u2220B=\u2220ACD
Again , \u2220A+\u2220B+\u2220C=\u2220ACD+\u2220C=180\u00b0 [\u2235 Line AC stands on ray BCD \u2220C+\u2220ACD=180\u00b0]<\/p>\n\n\n\n
\n\n\n\nQuestion 29<\/h4>\n\n\n\n
<fig to be added>
[Hint: \u2220BAL=\u2220CAL=\u22201 (say) [\u2235 AL is bisector of \u2220BAC]
In \u0394ABL , \u2220ALC=\u2220B+\u22201
or 2\u2220ALC=2\u2220B+2\u22201..(i)
In \u0394ABC, exterior \u2220ACD=\u2220B+\u2220A=\u2220B+2\u22201
\u2234 \u2220B+\u2220ACD=\u2220B+\u2220B+2\u22201=2\u2220B+2\u22201..(ii)
From (i) and (iii) , we get
2\u2220ALC=\u2220B+\u2220ACD=\u2220ABC+\u2220ACD ]<\/p>\n\n\n\n
\n\n\n\nQuestion 30<\/h4>\n\n\n\n
<fig to be added>
[Hint: We know that a regular pentagon has all its five sides equal. Also angle subtended by any side at the centre O=\\dfrac{360^{\\circ}}{5}=72^{\\circ}
\u2234 \u2220COD=72\u00b0
In \u0394OCD, OC=OD
\u2234 \u2220OCD=\u2220ODC=\\dfrac{180^{\\circ}-72^{\\circ}}{2}=54^{\\circ}
\u2234 \u2220OCF=180\u00b0-54\u00b0=126\u00b0
Consider \u0394AOB and \u0394AOE
AB=AE; \u2220BAO=\u2220EAO [\u2235 AO is bisector of \u2220BAE]
AO=AO
\u2234 \u0394AOB\u2245\u0394AOE; BO=OE
Now , AO will bisect \u2220COD
\u2234 \u2220COM=\u2220MOD=\\dfrac{72^{\\circ}}{2}=36^{\\circ}
\u2234 \u2220OMC=\u2220OCM+\u2220COM=54\u00b0+36\u00b0=90\u00b0
Hence, \u2220AMC=90\u00b0]<\/p>\n\n\n\n
\n\n\n\nQuestion 31<\/h4>\n\n\n\n
<fig to be added>
[Hint: AE is bisector of \u2220CAD
\u2234 \u2220CAE=\u2220DAE=\u22201 (say)
Now exterior \u2220CAD=\u2220B+\u2220C\u21d22\u22201
But \u2220B+\u2220C=2\u2220C [\u2235 \u2220B=\u2220C]
\u2234 2\u2220C=2\u22201 or \u2220C=\u22201=\u2220CAE
\u2235 These are alternate angles
\u2234 AE||BC ]<\/p>\n\n\n\n
\n\n\n\nQuestion 32<\/h4>\n\n\n\n
[Hint: In figure (i) , \u2220BFO=\u2220EDO=90\u00b0
\u2220BFO=\u2220EOD [Vertically opposite angles]
\u2234 \u2220BFO+\u2220BOF=\u2220EDO+\u2220EOD
or 180\u00b0-\u2220FBO=180\u00b0-\u2220OED
or \u2220FBO=\u2220OED
i.e., \u2220ABC=\u2220FED
In figure (ii) , \u2220B+\u2220D+\u2220E+\u2220F=360\u00b0
or \u2220B+90\u00b0+\u2220E+90\u00b0=360\u00b0
or \u2220B+\u2220E=180\u00b0 ]<\/p>\n\n\n\n
\n\n\n\nQuestion 33<\/h4>\n\n\n\n
Prove that \u2220P+\u2220Q$=\\dfrac{1}{2}(\\angle ABC+ \\angle ADC)$<\/strong>
[Hint: Let \u2220ABC=2x and \u2220ADC=2y, then in quadrilateral PBQD, we have
\u2220P+\u2220PDQ+\u2220Q+\u2220QBP=360\u00b0
or \u2220P+(180\u00b0-y)+\u2220Q+(180\u00b0-x)=360\u00b0
or \u2220P+\u2220Q=x+y$=\\dfrac{1}{2}(\\angle ABC+\\angle ADC)$
[(\u2235 2x=\u2220ABC \u2234 $x=\\dfrac{1}{2}\u2220ABC$ and 2y=\u2220ADC
\u2234$y=\\dfrac{1}{2}\\angle ADC$ )]<\/p>\n\n\n\n