{"id":566128,"date":"2021-12-29T09:52:15","date_gmt":"2021-12-29T09:52:15","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=566128"},"modified":"2021-12-29T12:00:55","modified_gmt":"2021-12-29T12:00:55","slug":"maharashtra-board-solutions-for-class-10-maths-part-2-chapter-6-trigonometry","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-for-class-10-maths-part-2-chapter-6-trigonometry\/","title":{"rendered":"Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 6- Trigonometry"},"content":{"rendered":"\n

Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 6- Trigonometry<\/h2>\n\n\n\n

Maharashtra Board 10th Maths Chapter 6, Class 10 Maths Chapter 6 solutions<\/p>\n\n\n\n

Practice Set 6.1<\/h4>\n\n\n\n

Question 1.
If sin \u03b8 = 725, find the values of cos \u03b8 and tan \u03b8.
Solution:<\/strong>
sin \u03b8 = 725 \u2026 [Given]
We know that,
sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1
\"Maharashtra
\u2026[Taking square root of both sides] Now, tan \u03b8 = sin\u03b8cos\u03b8
\"Maharashtra
Alternate Method:
sin \u03b8 = 725 \u2026(i) [Given]
Consider \u2206ABC, where \u2220ABC 90\u00b0 and \u2220ACB = \u03b8.
sin \u03b8 = ABAC \u2026 (ii) [By definition]
\u2234 ABAC = 725 \u2026 [From (i) and (ii)]
\"Maharashtra
LetAB = 7k and AC = 25k
In \u2206ABC, \u2220B = 90\u00b0
\u2234 AB2<\/sup> + BC2<\/sup> = AC2<\/sup> \u2026 [Pythagoras theorem]
\u2234 (7k)2<\/sup> + BC2<\/sup> = (25k)2<\/sup>
\u2234 49k2<\/sup> + BC2<\/sup> = 625k2<\/sup>
\u2234 BC2<\/sup> = 625k2<\/sup> \u2013 49k2<\/sup>
\u2234 BC2<\/sup> = 576k2<\/sup>
\u2234 BC = 24k \u2026[Taking square root of both sides]
\"Maharashtra<\/p>\n\n\n\n

Question 2.
If tan \u03b8 = 34, find the values of sec \u03b8 and cos \u03b8.
Solution:<\/strong>
\"Maharashtra
Alternate Method:
tan \u03b8 = 34 \u2026(i)[Given]
Consider \u2206ABC, where \u2220ABC 90\u00b0 and \u2220ACB = \u03b8.
tan \u03b8 = ABBC \u2026 (ii) [By definition]
\u2234 ABBC = 34 \u2026 [From (i) and (ii)]
\"Maharashtra
Let AB = 3k and BC 4k
In \u2206ABC,\u2220B = 90\u00b0
\u2234 AB2<\/sup> + BC2<\/sup> = AC2<\/sup> \u2026[Pythagoras theorem]
\u2234 (3k)2<\/sup> + (4k)2<\/sup> = AC2<\/sup>
\u2234 9k2<\/sup> + 16k2<\/sup> = AC2<\/sup>
\u2234 AC2<\/sup> = 25k2<\/sup>
\u2234 AC = 5k \u2026[Taking square root of both sides]
\"Maharashtra<\/p>\n\n\n\n

Question 3.
If cot \u03b8 = 409, find the values of cosec \u03b8 and sin \u03b8
Solution:<\/strong>
\"Maharashtra
..[Taking square root of both sides]
\"Maharashtra
Alternate Method:
cot \u03b8 = 409 \u2026.(i) [Given]
Consider \u2206ABC, where \u2220ABC = 90\u00b0 and
\u2220ACB = \u03b8
cot \u03b8 = BCAB \u2026(ii) [By defnition]
\u2234 BCAB = 409 \u2026.. [From (i) and (ii)]
Let BC = 40k and AB = 9k
\"Maharashtra
In \u2206ABC, \u2220B = 90\u00b0
\u2234 AB2<\/sup> + BC2<\/sup> = AC2<\/sup> \u2026 [Pythagoras theorem]
\u2234 (9k)2<\/sup> + (40k)2<\/sup> = AC2<\/sup>
\u2234 81k2<\/sup> + 1600k2<\/sup> = AC2<\/sup>
\u2234 AC2<\/sup> = 1681k2<\/sup>
\u2234 AC = 41k \u2026 [Taking square root of both sides]
\"Maharashtra<\/p>\n\n\n\n

Question 4.
If 5 sec \u03b8 \u2013 12 cosec \u03b8 = \u03b8, find the values of sec \u03b8, cos \u03b8 and sin \u03b8.
Solution:<\/strong>
5 sec \u03b8 \u2013 12 cosec \u03b8 = 0 \u2026[Given]
\u2234 5 sec \u03b8 = 12 cosec \u03b8
\"Maharashtra
\"Maharashtra<\/p>\n\n\n\n

Question 5.
If tan \u03b8 = 1, then find the value of
\"Maharashtra
Solution:<\/strong>
tan \u03b8 = 1 \u2026 [Given]
We know that, tan 45\u00b0 = 1
\u2234 tan \u03b8 = tan 45\u00b0
\u2234 \u03b8 = 45\u00b0
\"Maharashtra<\/p>\n\n\n\n

Question 6.
Prove that:<\/strong>
i. sin2\u03b8cos\u03b8+cos\u03b8=sec\u03b8
ii. cos2<\/sup> \u03b8 (1+ tan2<\/sup> \u03b8) = 1
iii. 1\u2212sin\u03b81+sin\u03b8\u2212\u2212\u2212\u2212\u2212\u221a=sec\u03b8\u2212tan\u03b8
iv. (sec \u03b8 \u2013 cos \u03b8) (cot \u03b8 + tan \u03b8) tan \u03b8. sec \u03b8
v. cot \u03b8 + tan \u03b8 cosec \u03b8. sec \u03b8
vi. 1sec\u03b8\u2212tan\u03b8=sec\u03b8+tan\u03b8
vii. sin4<\/sup> \u03b8 \u2013 cos4<\/sup> \u03b8 = 1 \u2013 2 cos2<\/sup> \u03b8
viii. sec\u03b8+tan\u03b8=cos\u03b81\u2212sin\u03b8
\"Maharashtra
Proof:
<\/strong>i. L.H.S. = sin2\u03b8cos\u03b8+cos\u03b8
\"Maharashtra<\/p>\n\n\n\n

ii. L.H.S. = cos2<\/sup> \u03b8(1 + tan2<\/sup> \u03b8)
= cos2<\/sup> \u03b8 sec2<\/sup> \u03b8 \u2026[\u2235 1 + tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]
\"Maharashtra
= 1
= R.H.S.
\u2234 cos2<\/sup> \u03b8 (1 + tan2<\/sup> \u03b8) = 1<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

iv. L.H.S. = (sec \u03b8 \u2013 cos \u03b8) (cot \u03b8 + tan \u03b8)
\"Maharashtra
\u2234 (sec \u03b8 \u2013 cos \u03b8) (cot \u03b8 + tan \u03b8) = tan \u03b8. sec \u03b8<\/p>\n\n\n\n

v. L.H.S. = cot \u03b8 + tan \u03b8
\"Maharashtra
\u2234 cot \u03b8 + tan \u03b8 = cosec \u03b8.sec \u03b8<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

vii. L.H.S. = sin4<\/sup> \u03b8 \u2013 cos4<\/sup> \u03b8
= (sin2<\/sup> \u03b8)2<\/sup> \u2013 (cos2<\/sup> \u03b8)2<\/sup>
= (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) (sin2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8)
= (1) (sin2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8) \u2026.[\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1]
= sin2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8
= (1 \u2013 cos2<\/sup> \u03b8) \u2013 cos2<\/sup> \u03b8 \u2026[\u03b8 sin2<\/sup> \u03b8 = 1 \u2013 cos2<\/sup> \u03b8]
= 1 \u2013 2 cos2<\/sup> \u03b8
= R.H.S.
\u2234 sin4<\/sup> \u03b8 \u2013 cos4<\/sup> \u03b8 = 1 \u2013 2 cos2<\/sup> \u03b8<\/p>\n\n\n\n

viii. L.H.S. = sec \u03b8 + tan \u03b8
\"Maharashtra<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

xi. L.H.S. = sec4<\/sup> A (1 \u2013 sin4<\/sup> A) \u2013 2 tan2<\/sup> A
= sec4<\/sup> A [12<\/sup> \u2013 (sin2<\/sup> A)2<\/sup>] \u2013 2 tan2<\/sup> A
= sec4<\/sup> A (1 \u2013 sin2<\/sup>A) (1 + sin2<\/sup> A) \u2013 2 tan2<\/sup> A
= sec4<\/sup> A cos2<\/sup>A (1 + sin2<\/sup> A) \u2013 2 tan2<\/sup>A
[ \u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1 ,\u2235 1 \u2013 sin2<\/sup> \u03b8 = cos2<\/sup> \u03b8]
\"Maharashtra
\"Maharashtra<\/p>\n\n\n\n

Question 1.
Fill in the blanks with reference to the figure given below. (Textbook pg. no. 124)
\"Maharashtra
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 2.
Complete the relations in ratios given below. (Textbook pg, no. 124)<\/strong>
\"Maharashtra
Solution:
<\/strong>i. sin\u03b8cos\u03b8 = [tan \u03b8]
ii. sin \u03b8 = cos (90 \u2013 \u03b8)
iii. cos \u03b8 = (90 \u2013 \u03b8)
iv. tan \u03b8 \u00d7 tan (90 \u2013 \u03b8) = 1<\/p>\n\n\n\n

Question 3.
Complete the equation. (Textbook pg. no, 124)
sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = [______]
Solution:
sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = [1]<\/p>\n\n\n\n

Question 4.
Write the values of the following trigonometric ratios. (Textbook pg. no. 124)<\/strong>
\"Maharashtra
Solution:
<\/strong>\"Maharashtra<\/p>\n\n\n\n

Practice Set 6.2<\/h4>\n\n\n\n

Question 1.
A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45\u00b0. Find the height of the church.
Solution:<\/strong>
Let AB represent the height of the church and point C represent the position of the person.
\"Maharashtra
BC = 80 m
Angle of elevation = \u2220ACB = 45\u00b0
In right angled \u2206ABC,
tan 45\u00b0 = ABBC \u2026 [By definition]
\u2234 1 = AB80
\u2234 AB = 80m
\u2234 The height of the church is 80 m.<\/p>\n\n\n\n

Question 2.
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60\u00b0. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. ( 3\u2013\u221a = 1.73)
Solution:<\/strong>
Let AB represent the height of lighthouse and point C represent the position of the ship.
AB = 90 m
\"Maharashtra
Angle of depression = \u2220PAC = 60\u00b0
Now, ray AP || seg BC
\u2234 \u2220ACB = \u2220PAC \u2026 [Alternate angles]
\u2234 \u2220ACB = 60\u00b0
In right angled \u2206ABC,
tan 60\u00b0 = ABBC \u2026 [By definition]
\"Maharashtra
\u2234 The ship is 51.90 m away from the lighthouse.<\/p>\n\n\n\n

Question 3.
Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60\u00b0. What is the height of the second building?
Solution:<\/strong>
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.
\"Maharashtra
Draw seg AM \u22a5 seg CD.
Angle of elevation = \u2220CAM = 60\u00b0
AB = 10 m
BD= 12 m
In \ua838ABDM,
\u2220B = \u2220D = 90\u00b0
\u2220M = 90\u00b0 \u2026 [seg AM \u22a5 seg CD]
\u2234 \u2220A = 90\u00b0 \u2026 [Remaining angle of \ua838ABDM]
\u2234 \ua838ABDM is a rectangle \u2026. [Each angle is 90\u00b0]
\u2234 AM = BD = 12 m opposite sides
DM = AB = 10 m of a rectangle
In right angled \u2206AMC,
tan 60\u00b0 = CMAM \u2026[By definition]
\u2234 3\u2013\u221a = CM12
\u2234 CM = 123\u2013\u221a m
Now, CD = DM + CM \u2026 [C \u2013 M \u2013 D]
\u2234 CD = (10 + 123\u2013\u221a)m
= 10 + 12 \u00d7 1.73
= 10 + 20.76 = 30.76
\u2234 The height of the second building is 30.76 m.<\/p>\n\n\n\n

Question 4.
Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops is 22 metre. Find the angle made by the wire with the horizontal.
Solution:<\/strong>
Let AB and CD represent the heights of two poles, and AC represent the length of the wire.
\"Maharashtra
Draw seg AM \u22a5 seg CD.
Angle of elevation = \u2220CAM = \u03b8
AB = 7 m
CD = 18 m
AC = 22 m
In \ua838ABDM,
\u2220B = \u2220D = 90\u00b0
\u2220M = 90\u00b0 \u2026[seg AM \u22a5 seg CD]
\u2234 \u2220A = 90\u00b0 \u2026 [Remaining angle of \ua838ABDM]
\u2234 \u25a1ABDM is a rectangle. \u2026 [Each angle is 90\u00b0]
\u2234 DM = AB = 7 m \u2026 [Opposite sides of a rectangle]
Now, CD = CM + DM \u2026 [C \u2013 M \u2013 D]
\u2234 18 = CM + 7
\u2234 CM = 18 \u2013 7 = 11 m
In right angled \u2206AMC,
sin \u03b8 = CMAC \u2026..[By definition]
\u2234 sin \u03b8 = 1122 = 12
But, sin 30\u00b0 = 12
\u2234 \u03b8 = 30\u00b0
\u2234 The angle made by the wire with the horizontal is 30\u00b0.<\/p>\n\n\n\n

Question 5.
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60\u00b0 with the horizontal. Find the height of the tree.
Solution:<\/strong>
Let AB represent the height of the tree.
Suppose the tree broke at point C and its top touches the ground at D.
AC is the broken part of the tree which takes position CD such that \u2220CDB = 60\u00b0
\u2234 AC = CD \u2026(i)
BD = 20 m
In right angled \u2206CBD,
\"Maharashtra
tan 60\u00b0 = BCBD \u2026 [By definition]
\u2234 3\u2013\u221a = BC20
\u2234 BC = 203\u2013\u221a m
Also, cos 60\u00b0 = BCCD \u2026 [By definition]
\u2234 12 = 20CD
\u2234 CD = 20 \u00d7 2 = 40 m
\u2234 AC = 40 m \u2026[From(i)]
Now, AB = AC + BC \u2026.[A \u2013 C \u2013 B]
= 40 + 203\u2013\u221a
= 40 + 20 \u00d7 1.73
= 40 + 34.6
= 74.6
\u2234 The height of the tree is 74.6 m.<\/p>\n\n\n\n

Question 6.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60\u00b0 with the ground. Assuming that the string is straight, find the length of the string. (3\u2013\u221a = 1.73)
Solution:<\/strong>
Let AB represent the height at which kite is flying and point C represent the point where the string is tied at the ground.
\u2220ACB is the angle made by the string with the ground.
\"Maharashtra
\u2220ACB = 60\u00b0
AB = 60 m
In right angled \u2206ABC,
sin 60\u00b0 = ABAC \u2026 [By definition]
\"Maharashtra
\u2234 AC = 40 3\u2013\u221a = 40 \u00d7 1.73 = 69.20 m
\u2234 The length of the string is 69.20 m.<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 6- Trigonometry<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 6- Trigonometry PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 2) :<\/strong><\/h2>\n\n\n\n