Question

The angle of depression from the top of one building to the foot of a building across the street is 53°. The angle of depression to the top of the same building is 19°. The two buildings are 80 feet apart. What is the height of the shorter building?

Answer 1

Answer:

78.64 feet

Step-by-step explanation:

Refer the attached figure

Height of the tall building = AC

The angle of depression from the top of one building to the foot of a building across the street is 53°. i.e.∠ADC = 53°

The angle of depression to the top of the same building is 19°.i.e.∠AEB = 19°

Height of small building = ED

The two buildings are 80 feet apart i.e. CD =BE= 80 feet

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 53^{\circ}= \frac{AC}{CD}$

$Tan 53^{\circ}= \frac{AC}{80}$

$1.327 \times 80= AC$

$106.16= AC$

In ΔABE

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 19^{\circ}= \frac{AB}{BE}$

$Tan 19^{\circ}= \frac{AB}{80}$

$0.344 \times 80=AB$

$27.52= AB$

AC - AB = BC

106.16-27.52=BC

78.64=BC

So, BC = ED = 78.64 feet

Hence the height of the shorter building is 78.64 feet