Question 1

From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30^ and 45^ . The height of tower is

Accepted Answer

Accoding to condition, Let, height of tower =h In A B D, 45^ = A B B D B D=50 ~m Now, In A C C^ aligned & & 30^ & = A C^ C^ C \ & & 1 3 & = 50-h 50 \ & & 50 & =50 3 -h 3 \ & & h 3 & =50( 3 -1) \ & & h & =50 (1- 1 3 ) \ & & & =50 (1- 3 3 ) m aligned

Question 2

A B C is a triangular park with A B=A C=100 ~m . A TV tower stands at the mid-point of B C. The angles of elevation of the top of the tower at A, B, C are 45^ , 60^ , 60^ respectively. The height of t

Accepted Answer

According to question, Let D E=h and C D=D B=x In E B D, 60^ = h x x= h 3 Now, in A D E aligned & 45^ = E D D A \ & D A=h aligned In A B D, by applying Pythagoras theorem aligned & ( h 3 )^2+h^2=100^2 \ & 4 h^2 3 =10000 h=50 3 aligned

Question 3

Let be the solution of 16^ ^2 +16^ ^2 =10 in (0, 4 ). If the shadow of a vertical pole is 1 3 of its height, then the altitude of the sun is

Accepted Answer

We have, 16^ ^2 +16^ ^2 =10 16^ ^2 +16^ 1- ^2 =10 x+ 16 x =10, where x=16^ ^2 x^2-10 x+16=0 x=2,8 16^ ^2 =2,8 2^ 4 ^2 =2,2^3 4 ^2 =1,3 ^2 = 1 4 , ( 3 2 )^2 = 1 2 , 3 2 = 6 , 3 Let the altitude of the sun be . Then, aligned & & & = h h / 3 = 3 \ & & & = / 3 =2 aligned

Question 4

A house subtends a right angle at the window of a opposite house and the angle of elevation of the window from the bottom of the first house is 60. If the distance between two houses be 6 ~m , then th

Accepted Answer

Let P Q be the house subtending a right angle at the window B of opposite house A B In P A B, we have 60^ = A B 6 =A B=6 3 ~m In C B Q, we have 30^ = h-C P B C array ll & 1 3 = h-6 3 6 [ A B=C P, B C=A P] \ & h=6 ( 3 + 1 3 ) \ & h=8 3 ~m array

Question 5

ABC is a triangular park with AB = AC =100 ~m . A TV tower stands at the mid-point of BC . The angles of elevation of the top of the tower at A, B, C are 45^ , 60^ , 60^ respectively. The height of th

Accepted Answer

45^ = PQ AQ = h AQ h = AQ Where PQ is tower and ABC is the park, with Q being mid point of the side BC and PQ = h Also, A Q^ 2 +B Q^ 2 =100^ 2 h ^ 2 + h ^ 2 ^ 2 60^ =100^ 2 h ^ 2 [1+ 1 3 ]=100^ 2 h ^ 2 = 3 100^ 2 4 h =50 3

Heights and Distances — Previous Year Questions & Solutions