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Chapter 2 Class 12 Inverse Trigonometric Functions
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Two men on either side of a temple of 30 meters high observe its top at the angles of elevation 𝛼 and 𝛽 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 = 𝛼 =

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

Β 

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Question 2

∠CAB = 𝛼 =

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

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Question 3

∠BCA = β =

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

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Question 4

∠ABC =

(a) Ο€/4
(b) Ο€/6
(c) Ο€/2
(d) Ο€/3

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Question 5

Domain and Range of cos βˆ’1 x =

(a) (βˆ’1, 1), (0 , πœ‹)
(b) [βˆ’1, 1], (0 , πœ‹)
(c) [βˆ’1, 1], [0 , πœ‹]
(d) (βˆ’1, 1), [-Ο€/2, Ο€/2]

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Transcript

Two men on either side of a temple of 30 meters high observe its top at the angles of elevation 𝛼 and 𝛽 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: Question 1 ∠CAB = 𝛼 = (a) 〖𝑠𝑖𝑛〗^(βˆ’1) (2/√3) (b) 〖𝑠𝑖𝑛〗^(βˆ’1) (1/2) (c) 〖𝑠𝑖𝑛〗^(βˆ’1) (2) (d) 〖𝑠𝑖𝑛〗^(βˆ’1) (√3/2) In Ξ” ABD tan 𝜢 = 𝐡𝐷/𝐴𝐷 tan 𝜢 = 30/(30√3) tan 𝜢 = 1/√3 ∴ 𝜢 = 30Β° = 𝝅/πŸ” Thus, sin 𝜢 = sin 30Β° = 1/2 So, sin 𝜢 = 1/2 𝜢 = γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) (𝟏/𝟐) So, correct answer is (b) Question 2 ∠CAB = 𝛼 = (a) γ€–π‘π‘œπ‘ γ€—^(βˆ’1) (1/5) (b) γ€–π‘π‘œπ‘ γ€—^(βˆ’1) (2/5) (c) γ€–π‘π‘œπ‘ γ€—^(βˆ’1) (√3/2) (d) γ€–π‘π‘œπ‘ γ€—^(βˆ’1) (4/5) Now, 𝜢 = 30Β° = 𝝅/πŸ” Thus, cos 𝜢 = cos 30Β° = √3/2 So, sin 𝜢 =√3/2 𝜢 = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) (√3/2) So, correct answer is (c) Question 3 ∠BCA = 𝛽 = (a) γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (1/2) (b) γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (2) (c) γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (1/√3) (d) γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (√3) In Ξ” ABD tan 𝛽 = 𝐡𝐷/𝐴𝐷 tan 𝛽 = 30/(10√3) tan 𝛽 = 3/√3 tan 𝛽 = √3 𝛽 = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (√3) So, the correct answer is (d) Also, 𝛽 = 60Β° = 𝝅/πŸ‘ Question 4 ∠ABC = (a) πœ‹/4 (b) πœ‹/6 (c) πœ‹/2 (d) πœ‹/3 Since 𝜢 = 30Β° and 𝛽 = 60Β° In Ξ” ABC, By Angle sum property 𝜢 + 𝛽 + ∠ ABC = 180Β° 30Β° + 60Β° + ∠ ABC = 180Β° ∠ ABC = 90Β° ∠ ABC = 𝝅/𝟐 So, the correct answer is (c) Question 5 Domain and Range of cosβˆ’1 π‘₯ = (a) (βˆ’1, 1), (0 , πœ‹) (b) [βˆ’1, 1], (0 , πœ‹) (c) [βˆ’1, 1], [0 , πœ‹] (d) (βˆ’1, 1), [βˆ’πœ‹/2, πœ‹/2] Since cos x is defined at x = 0, and x = πœ‹ Domain of cosβˆ’1 π‘₯ includes βˆ’1 and 1 Range of cosβˆ’1 π‘₯ also includes 0 and πœ‹ So, the correct answer is (c)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.