Question

what is the height of the building if it is 38cm away from the observer and its angle of elevation is 30 degree?

Answer 1

let the height of the building be " h cm "

given , distance between the observer
and the building BC = 38 cm

.
angle of elevation = 30°

.
since ,we know when we have given base and we have to find the height of building etc.
then , we solve it either for tanA or cotA .

both gives us the same same value.

.
so let us solve it for tanA,

.
in ∆ ABC :
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tan 30° = height / base

.
tan30° = AB/ BC

.
1/ √3 = AB/ 38

.
1/ √3 = h/ 38

.
h = 38/ √3 = ( 38/√3 ) × ( √3/ √3 )

.
h = 38√3/ 3 = 12.67 × √3

.
h =21. 9 cm

Answer:
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height of the building = 21.9cm

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