Question

From the top of a building 60m high the angular depression of the top ofa pyramid is 45
and angular depression of the foot of the pyramid is 60" Find the height of the pyramid

Answer 1

Answer:

Height of the pyramid is 25.36 m

Step-by-step explanation:

Let AB be the building of height 60 m.

Let h be the height of the pyramid.

In ΔAA'C,

$tan\:45=\frac{AA'}{A'C}\\\\1=\frac{60-h}{A'C}\\\\A'C = 60 - h$

From the figure,

BD=A'C= 60-h

In ΔABD,

$tan\:60=\frac{AB}{BD}\\\\\sqrt{3}=\frac{60}{60-h}\\\\60-h=\frac{60}{\sqrt{3}}\\\\60-h=20\sqrt{3}\\\\60-20\sqrt{3}=h\\\\h=60-(20*1.732)\\\\h=60-34.64\\\\h= 25.36\:meters$

Answer 2

Answer:

Height of pyramid is 25.36 m

Step-by-step explanation:

Given: Height of Building, AB = 60 m

           Angle of depression of the top of pyramid = 45°

          Angle of depression of the foot of the pyramid = 60°

To Find: Height of Pyramid, CD

Height of Pyramid, CD = EB = AB - AE

Angle of depression of top of pyramid is equal to angle of elevation from top of pyramid to top of building. Also, angle of depression of foot of pyramid is equal to angle of elevation from foot of pyramid to top of building.

⇒ m∠EDA = 45°  and m∠ACB = 60°

So using trigonometric ratios in ΔABC  and ΔADE , we get

from ΔABC

$tan\,60\,=\,\frac{AB}{CB}$

$\sqrt{3}\,=\,\frac{60}{CB}$

$CB\,=\,\frac{60}{\sqrt{3}}$

use this in ΔADE

$tan\,45\,=\,\frac{AE}{DE}$

$1\,=\,\frac{AE}{CB}$    (∵ from figure, DE = CB)

$1\,=\,\frac{AE}{\frac{60}{\sqrt{3}}}$

$1\,\times\frac{60}{\sqrt{3}}=\,AE$

$AE\,=\,\frac{60}{\sqrt{3}}$

$AE\,=\,\frac{60}{1.73}$

AE = 34.64

⇒ CD = BE = 60 - 34.64 =  25.358 = 25.36 (approx.)

Therefore, Height of pyramid is 25.36 m