Question

The angle of elevation of a tower $CD$ at a place $A$ due to south of it is $60$ and at a place $B$ due west of $A$, the elevation is $30$. if $AB=3km$, then find the height of tower.

• A. $0.95km$
• B. $200m$
• C. $173.2m$
• D. $141.4m$

Answer 1

Answer: (A)

$0.95km$

We have,

$CDasthegiventower.$

$Let\angle DAC={60}^0$

$Leth=heightoftower=CD$

$In\Delta ADC,\cot {60}^0=\frac{DA}{CD}\Rightarrow \frac{1}{\sqrt{3}}=\frac{DA}{h}\Rightarrow DA=\frac{h}{\sqrt{3}}.......\left(1\right)In\Delta BDC,\cot {30}^0=\frac{BD}{CD}\Rightarrow \sqrt{3}=\frac{BD}{h}.........\left(2\right)$

$In\Delta ADB,A{B}^2=A{D}^2+B{D}^2{\left(3\right)}^2={\left(\frac{h}{\sqrt{3}}\right)}^2+{\left(\sqrt{3}h\right)}^2\Rightarrow 9=\left(\frac{h^2}{3}\right)+{\left(\sqrt{3}h\right)}^2\Rightarrow 10h^2=9\Rightarrow h^2=\frac{9}{10}\Rightarrow h=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}=\frac{3\times 3.162}{10}=0.9486km=0.95km$

Hence, this is the answer.