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The angle of elevation of the …

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The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°. The angle of elevation of the top of second tower from the foot of the first tower is 30°. Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.
  • 5 answers

Preeti Dabral 1 year ago

Let distance between the two towers = AB = x m
and height of the other tower = PA = h m
Given that, height of the tower = QB = 30 m and {tex}\angle{/tex}QAB = 60°, {tex}\angle{/tex}PBA = 30°.

Now, in {tex}\Delta Q A B, \tan 60^{\circ}=\frac{Q B}{A B}=\frac{30}{x}{/tex}
{tex}\Rightarrow \sqrt{3}=\frac{30}{x}{/tex}
{tex}x=\frac{x^{2}}{3}-\frac{x}{3}-\frac{x-3}{3}-x+0.5 \pi{/tex}
and in {tex}\Delta P B A{/tex}
{tex}\tan 30^{\circ}=\frac{P A}{A B}=\frac{h}{x}{/tex}
{tex}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{10 \sqrt{3}}{/tex} {tex}[\because x=10 \sqrt{3} \mathrm{m}]{/tex}
{tex}\Rightarrow h = 10 m{/tex}
Hence, the requied distance and height are {tex}10\sqrt3{/tex} and 10m, respectively.

Ankit Class 1 year ago

Ye figure draw kaise ki ho batao

Ankit Class 1 year ago

PQ is the length of the wire attached to the tops of both the towers

Ankit Class 1 year ago

PQ=√x^2 +(30-h)^2 . PQ=√(10√3)^2 + (30-10)^2 . PQ=√ 300+400. PQ=√700. PQ= 10√7 m

Ankit Class 1 year ago

Math standard mein ye question puchha tha n mere mein bhi puchha tha set 1 tha mera 21 feb ko paper tha n
http://mycbseguide.com/examin8/

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