The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For the viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following: 1. Measure of ∠ = a. tan−1(2) b. tan−1( 12) c. tan−1( 1) d. tan−1( 3) 2. ∠ = a. tan−1(34) b. tan−1(3) c. tan−1(43) d. tan−1(4) 3. ∠= a. tan−1(11) b. tan−13 c. tan−1(211) d. tan−1(112) 4. | Is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠ and ∠′ Is a. tan-1(1/2) b. tan-1 (1/8) c. tan−1( 25) d. tan−1(1121) 5. Domain and Range of tan−1= a. +, (−2,2) b. −, (−2,2) c. R , (−2,2) d. R , (0 ,2)
Answers
1. Measure of ∠CAB = ?
a. tan^−1(2) b. tan^−1(1/2) c. tan^−1(1) d. tan^−1(3) ?
Answer :-
we know that,
→ tan θ = Perpendicular / Base .
→ tan ∠CAB = CB / AB
→ tan ∠CAB = 10/20
→ ∠CAB = tan^(-1)(1/2) (b) (Ans.)
2. ∠DAB = ?
a. tan−1(3/4) b. tan−1(3) c. tan−1(4/3) d. tan−1(4)
Answer :-
given that,
→ ∠DAB = 2*∠CAB
so,
→ tan ∠DAB = tan2(∠CAB)
→ tan ∠DAB = 2tan(∠CAB) / (1 - tan²∠CAB) = 2(1/2)/[1 - (1/2)²] = 1 / (1 - 1/4) = 1 / (3/4) = (4/3)
then,
→ ∠DAB = tan^(-1)(4/3) (C) (Ans.)
3. ∠EAB = ?
a. tan−1(11) b. tan−13 c. tan−1(2/11) d. tan−1(11/2)
Answer :-
given that,
→ ∠EAB = 3*∠CAB
so,
→ tan ∠EAB = tan3(∠CAB)
→ tan ∠EAB = (3tanA - tan³A)/(1 - 3tan²A) = [3(1/2) - (1/2)³] / [1 - 3(1/2)²] = [(3/2) - (1/8)] / [1 - (3/4)] = (11/8) / (1/4) = (11/2)
then,
→ ∠EAB = tan^(-1)(11/2) (D) (Ans.)
4. A' Is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠CAB and ∠CA′B Is ?
a. tan-1(1/2) b. tan-1 (1/8) c. tan−1( 25) d. tan−1(1121)
Answer :-
→ ∠CAB - ∠CA′B = tan^(-1)(1/8)
5. Domain and Range of tan−1 x = ?
a. +, (−2,2) b. −, (−2,2) c. R , (−2,2) d. R , (0 ,2)
Answer :-
→ Domain and Range of tan^(−1) x = R, (-π/2 , π/2) .
