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Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.