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  • On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower

On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.

 

 

 

 
 
 
 
 

Answers (1)

Let the height of the tower (AB) be h.

\angle AD\!B=\theta

\angle AC\!B=90- \theta    (angles are complementary)

Now, in  \bigtriangleup ABC, \angle B =90^{\circ}\tan (90-\theta )=\frac{AB}{BC}=\frac{h}{4}\; \: \: \: \: \: \: \; \; \; (\because BC=4\: m)

\cot \theta=\frac{h}{4}\: \: \: \: \: \: -(i)

In  \bigtriangleup AB\!D,

\tan \theta =\frac{AB}{BD}=\frac{h}{16}\; \; \; \; \; \; \; (\because BD=16\: m)

\tan \theta =\frac{h}{16}

\frac{1}{\cot \theta } =\frac{h}{16} \; \; \; \; \; \; \left ( \because \tan \theta = \frac{1}{\cot \theta} \right )

\cot \theta =\frac{16}{h}

\Rightarrow \frac{h}{4}=\frac{16}{h} \; \; \; \; (\text{from - (i)})

\Rightarrow h^2=4 \times 16

\Rightarrow h= \sqrt{64}=8\: m

\Rightarrow h=8\: m

 

 

 

 

Posted by

Ravindra Pindel

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