Question

From the top of a vertical cliff 50 m high, the angle of depression of an object that is level with the base of the cliff is 70°. How far is the object from the base of the cliff?

Answer 1

Answer:

The distance of the object from the base of cliff is 18.24 meters

Step-by-step explanation:

Given as :

The height of the vertical cliff = h = 50 meters

The distance of the object from the base of cliff = x meters

Let The angle of depression of object that level with cliff base = Ф = 70°

Now, from figure

Tan angle = $\dfrac{\textrm perpendicular}{\textrm base}$

I.e TanФ = $\dfrac{\textrm AB}{\textrm OA}$

Or, TanФ = $\dfrac{\textrm h}{\textrm x}$

Or, Tan 70° = $\dfrac{\textrm 50 meters}{\textrm x meters}$

Or, 2.74 = $\dfrac{\textrm 50}{\textrm x}$

∴ x =  $\dfrac{\textrm 50}{\textrm 2.74}$

I,e x = 18.24 meters

So, The distance of the object from the base of cliff = x = 18.24 meters

Hence,The distance of the object from the base of cliff is 18.24 meters Answer