1.

On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is​

Answer»

\huge\underline\mathfrak\pink{Answer}

LET AS be the TOWER and BD be the ground.

Let H be the HEIGHT of the tower.

In ΔADB,

\sf \: tan \: 30 \degree = \frac{h}{20+x} \\ \\ \sf \implies \frac{1}{ \sqrt{3} } = \frac{h}{20+x} \\ \\ \sf \implies \: h= \frac{20+x}{ \sqrt{3} } \\

In ΔADB,

\sf \implies \: tan \: 60 \degree = \frac{h}{x} \\ \\ \sf \implies\: h= \sqrt{3} x \\ \\ \sf \: Now, \sqrt{3} = \frac{20+x}{ \sqrt{3} } \\ \\ \sf \implies \: 3x=20+x \\ \\\sf \implies \: x=10 \\ \\ \sf \implies \: h= \sqrt{3} ×10=10 \sqrt{3} \: m


Discussion

No Comment Found

Related InterviewSolutions