Two men on either side of a temple of 30 meters high observe its top at the angles of elevation ๐›ผ and ๐›ฝ respectively. (as shown in the figure above). The distance between the two men is 40√3 meters and the distance between the first person A and the temple is 30 √3 meters. Based on the above information answer the following:

Two Men on Either Side of a Temple - Teachoo.jpg

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Question 1

∠CAB = ๐›ผ =

(a)  sin -1   (2/√3)
(b)  sin -1   (1/2)
(c)  sin -1   (2)
(d)  sin -1  (√3/2)

 

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Question 2

∠CAB = ๐›ผ =

(a)  cos -1   (1/5)
(b)  cos -1   (2/5)
(c)  cos -1   (√3/2)
(d)  cos -1   (4/5)

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Question 3

∠BCA = β =

(a)  tan (-1)   (1/2)
(b)  tan (-1)  (2)
(c)  tan (-1)   (1/√3)
(d)  tan (-1)  (√3)

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Question 4

∠ABC =

(a) π/4
(b) π/6
(c) π/2
(d) π/3

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Question 5

Domain and Range of cos −1 x =

(a) (−1, 1), (0 , ๐œ‹)
(b) [−1, 1], (0 , ๐œ‹)
(c) [−1, 1], [0 , ๐œ‹]
(d) (−1, 1), [-π/2, π/2]

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  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise

Transcript

Two men on either side of a temple of 30 meters high observe its top at the angles of elevation ๐›ผ and ๐›ฝ respectively. (as shown in the figure above). The distance between the two men is 40โˆš3 meters and the distance between the first person A and the temple is 30 โˆš3 meters. Based on the above information answer the following: Question 1 โˆ CAB = ๐›ผ = (a) ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (2/โˆš3) (b) ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (1/2) (c) ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (2) (d) ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (โˆš3/2) In ฮ” ABD tan ๐œถ = ๐ต๐ท/๐ด๐ท tan ๐œถ = 30/(30โˆš3) tan ๐œถ = 1/โˆš3 โˆด ๐œถ = 30ยฐ = ๐…/๐Ÿ” Thus, sin ๐œถ = sin 30ยฐ = 1/2 So, sin ๐œถ = 1/2 ๐œถ = ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) (๐Ÿ/๐Ÿ) So, correct answer is (b) Question 2 โˆ CAB = ๐›ผ = (a) ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) (1/5) (b) ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) (2/5) (c) ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) (โˆš3/2) (d) ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) (4/5) Now, ๐œถ = 30ยฐ = ๐…/๐Ÿ” Thus, cos ๐œถ = cos 30ยฐ = โˆš3/2 So, sin ๐œถ =โˆš3/2 ๐œถ = ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) (โˆš3/2) So, correct answer is (c) Question 3 โˆ BCA = ๐›ฝ = (a) ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (1/2) (b) ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (2) (c) ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (1/โˆš3) (d) ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (โˆš3) In ฮ” ABD tan ๐›ฝ = ๐ต๐ท/๐ด๐ท tan ๐›ฝ = 30/(10โˆš3) tan ๐›ฝ = 3/โˆš3 tan ๐›ฝ = โˆš3 ๐›ฝ = ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (โˆš3) So, the correct answer is (d) Also, ๐›ฝ = 60ยฐ = ๐…/๐Ÿ‘ Question 4 โˆ ABC = (a) ๐œ‹/4 (b) ๐œ‹/6 (c) ๐œ‹/2 (d) ๐œ‹/3 Since ๐œถ = 30ยฐ and ๐›ฝ = 60ยฐ In ฮ” ABC, By Angle sum property ๐œถ + ๐›ฝ + โˆ  ABC = 180ยฐ 30ยฐ + 60ยฐ + โˆ  ABC = 180ยฐ โˆ  ABC = 90ยฐ โˆ  ABC = ๐…/๐Ÿ So, the correct answer is (c) Question 5 Domain and Range of cosโˆ’1 ๐‘ฅ = (a) (โˆ’1, 1), (0 , ๐œ‹) (b) [โˆ’1, 1], (0 , ๐œ‹) (c) [โˆ’1, 1], [0 , ๐œ‹] (d) (โˆ’1, 1), [โˆ’๐œ‹/2, ๐œ‹/2] Since cos x is defined at x = 0, and x = ๐œ‹ Domain of cosโˆ’1 ๐‘ฅ includes โˆ’1 and 1 Range of cosโˆ’1 ๐‘ฅ also includes 0 and ๐œ‹ So, the correct answer is (c)

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.