Question

The vertical cliff is 36 meters high. the angle of depression of an object which aligned with the base of the cliff is 28 degrees. find the distance of the object from the base of the cliff???

Answer 1

`67.67` metres.

Explanation:

In the picture I've drawn, `AB` is the Vertical Cliff. Let's Say the Point `C` is the position of the object aligned with the base of the cliff. `AD` || `BC`, and `/_CAD = 28^@` is the angle of depression.

As, `AD` || `BC`, and `AC` is the transversal,

`/_ CAD = /_ ACB = 28^@` [Alternate Angles]

Now, In Right `DeltaABC`,

`color(white)(xxx)tan` `angleACB = (AB)/(BC)`

`rArr tan 28^@ = 36/(BC)`

`rArr BC = 36/(tan 28^@ )= 36/0.532 = 67.67` metres (approx.)

So, The Distance of the Object from the Base of The Cliff

= `BC`

= `67.67` metres.