From the top and foot of a tower 40 m high, the angle of elevation of the top of a light house is found to be 30° and 60° respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower.

Let AC be the Tower such that AC = 40 m and BE be tine light house. Let CD be the horizontal from C. It is given that angles of elevation of the top of the light house from top and foot of the tower be 30° and 60° respectively.
i.e., ∠DCE = 30° and ∠BAE = 60°
Let AB = CD = x m and DE = x m
Now, in right triangle CDE, we have tan 30°


In right triangle ABE, we have


Comparing (i) and (ii), we gel



Hence,heighl of light house = BD + DE = 40 + 20 = 60 m.

In right triangle ABE, we have sin 60  

Hence, the distance of the top of the light house from the foot of the tower is 

 Let AB be the first pole such that AB = 60 m. Let the weight of the second pole be n metre. It is given that the angles of depression of the top A and the bottom B of the pole AB are 30° and 60° respectively.
∴ ∠ACE = 30° and ∠ADB = 60°
Let    BD = EC = x
Now, in right triangle ACE, we have


In right triangle ADB, we have

⇒    x = 20 × 1.732 = 34.64 m
Hence, width of the river = 34.64 m.
Comparing (i) and (ii), we gel



Hence,height of other pole = 40 m.

Let hetght of the pedestal BD be h metres, and angle of elevation of C and D at a point A on the ground be 60° and 45° respectively.

It is also given that the height of tine statue CD be 1.6 m
i.e.,    ∠CAB = 60°,
∠DAB = 45° and CD = 1.6m
In right triangle ABD, we have


In right triangle ABC, we have



Comparing (i) and (ii), we get

Hence, the height of pedestal
= 0.73 (1.73 + 1) = 0.73 × 2.73
= 1.929 m. = 2 m. (approx)

Let PC denote the multistroyed building and AB denotes the 8 m tall building.

We have ∠PBD = 30° and ∠PAC = 45°
Let    PD = h and AC = BD = x
Now, in  BPD, tan 30⁰ = 

Comparing (i) and (ii), we gel


Now, total height of mullistroycd building


And distance between two buildings(x)

Let A be the point of observation, C and E be the two points of the plane. It is given that after 15 seconds angle of elevation changes from 60° to 30°.


i.e., ∠BAC = 60° and ∠DAE = 30°. It is also given that height of the jet plane is  


[Sincc jet plane is flying at constant height, Let, CB = ED = h km]
In right triangle ABC, we have


In right triangle ADL, we have 


Comparing (i) and (ii) we get,


[Hence, constant height at which the jet is flying = 2.6 km (app)]

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