Let CE be the leaning tower. Let A and B be two given stations at distances a and b respectively from the foot of the tower.
Let CD = x and DE = h
In right triangle CDE, we have




In right triangle BDE, we have


In right triangle ADE, we have


Comparing (i) and (ii), we get



Comparing (i) and (iii), we get



Comparing (iv) and (v), we get




Hence, inclination ө to the horizontal is given by cot 

Let A be the boy and F is the position of kite. Let BF be vertical height of the kite and AF is the length of the string i.e. AF = 150 m. It is given that ∠BAF = 30°.


In right triangle ABF, we have

Let D be the position of second boy and DF be the length of second kite. It is given ∠EDF = 45°.
In right triangle DEF, we have


Hence, the length of the string = 

Let CE be a cliff of height h m. Angle of elevation of cliff from a fixed point A be ө.

i.e., ∠CAE = ө. ∠CAF = φ and AF = k metres. From F draw FD and FB perpendiculars on CE and AC respectively. It is also given that ∠DFE = α.
In right triangle ABF, we have



In right triangle ACE, we have

And,    DE = CE - CD = CE - BF
= h - k sin φ.
In right triangle DFE, we have





Hence, the height of the cliff is

Let AD be the lighthouse of height h m B and C are the position of the two ships which are on the either side of the road. The angles of depression of the ships from the top of the light house be 60° and 45°, ie., ∠ABD = 60 and ∠ACD = 45°

Let BD = x m and CD = y m.
In right triangle ABD, we have


In right triangle ACD, we have


Adding (i) and (ii), we get


Hence, the height of the light house = 200 m.

Let BC be the height of the tower and CD be the pole of height 5m fixed on the top of the tower. Let BC = h m.

The angle of elevation of top of the pole from point A on the ground be 60° and the angle of depression of the point A from the top of the tower be 45°, i.e. ∠ BAD = 60° and ∠ BAC = 45°.
In right triangle ABC, we have

In right triangle ABD, we have

Comparing (i) and (ii), we get