Case Based Questions (MCQ)

Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise

Β

## (a) (β1, 1), (0 , π) (b) [β1, 1], (0 , π) (c) [β1, 1], [0 , π] (d) (β1, 1), [-Ο/2, Ο/2]

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Two men on either side of a temple of 30 meters high observe its top at the angles of elevation πΌ and π½ respectively. (as shown in the figure above). The distance between the two men is 40β3 meters and the distance between the first person A and the temple is 30 β3 meters. Based on the above information answer the following: Question 1 β CAB = πΌ = (a) γπ ππγ^(β1) (2/β3) (b) γπ ππγ^(β1) (1/2) (c) γπ ππγ^(β1) (2) (d) γπ ππγ^(β1) (β3/2) In Ξ ABD tan πΆ = π΅π·/π΄π· tan πΆ = 30/(30β3) tan πΆ = 1/β3 β΄ πΆ = 30Β° = π/π Thus, sin πΆ = sin 30Β° = 1/2 So, sin πΆ = 1/2 πΆ = γπππγ^(βπ) (π/π) So, correct answer is (b) Question 2 β CAB = πΌ = (a) γπππ γ^(β1) (1/5) (b) γπππ γ^(β1) (2/5) (c) γπππ γ^(β1) (β3/2) (d) γπππ γ^(β1) (4/5) Now, πΆ = 30Β° = π/π Thus, cos πΆ = cos 30Β° = β3/2 So, sin πΆ =β3/2 πΆ = γπππ γ^(β1) (β3/2) So, correct answer is (c) Question 3 β BCA = π½ = (a) γπ‘ππγ^(β1) (1/2) (b) γπ‘ππγ^(β1) (2) (c) γπ‘ππγ^(β1) (1/β3) (d) γπ‘ππγ^(β1) (β3) In Ξ ABD tan π½ = π΅π·/π΄π· tan π½ = 30/(10β3) tan π½ = 3/β3 tan π½ = β3 π½ = γπ‘ππγ^(β1) (β3) So, the correct answer is (d) Also, π½ = 60Β° = π/π Question 4 β ABC = (a) π/4 (b) π/6 (c) π/2 (d) π/3 Since πΆ = 30Β° and π½ = 60Β° In Ξ ABC, By Angle sum property πΆ + π½ + β  ABC = 180Β° 30Β° + 60Β° + β  ABC = 180Β° β  ABC = 90Β° β  ABC = π/π So, the correct answer is (c) Question 5 Domain and Range of cosβ1 π₯ = (a) (β1, 1), (0 , π) (b) [β1, 1], (0 , π) (c) [β1, 1], [0 , π] (d) (β1, 1), [βπ/2, π/2] Since cos x is defined at x = 0, and x = π Domain of cosβ1 π₯ includes β1 and 1 Range of cosβ1 π₯ also includes 0 and π So, the correct answer is (c)