Question

. Two buildings of different heights stand 10 m apart. From the foot of the taller

building the top of the shorter one is seen at an angle of elevation of 45

and from

the foot of the shorter building the top of the taller building is seen at an elevation

of 70

.

a) Draw a rough figure .

b) what is the height of the shorter building.

c) What is height of the taller building.

(sin 70 =0.94,cos 70=0.34, tan 70=2.7)​

Answer 1

• a) Part is in the attachment
• Let the height of Shorter BUILDING BE h

according to the question

When seen from foot of shorter building the top of shorter building appears to be at angle of 45⁰

Hence A right angled triangle is formed with

BASE=10m  

PERPENDICULAR =h

HYPOTENUESE = ?(unknown)

We know that

         $tan(\theta)=\frac{pependicular}{base} \\ALSO \:THAT \:\\tan(45)=1\\As\: \theta=45\degree$

$hence\:\\tan(45)=\frac{h}{10} \\1=\frac{h}{10}\\ h=10m \\\bold{height\:of\:shorter\:building\:=10m}$

• Also when we watch from foot of taller building the angle of elevation appears to be 70° and A RIGHT ANGLED TRIANGLE IS FORMED

We know that

Perpendicular=H

Base=10

Hypotenuese=?(unknown)

$tan(\theta)=\frac{pependicular}{base}\\ \theta=70 \\tan(70)=\frac{H}{10}\\2.7=H/10\\H=27m\\\bold{height \:of\:taller\:building=27m}$