Math, asked by ismaeldalvi69, 6 months ago

The angle of elevation of the top of a tower 45m high, from the foot of another tower
in the same plane is 60 degree and the angle of elevation of the top of the second tower from the foot of the first tower is 30 degree . Find the distance between the two towers and also the height of the other tower.

Answers

Answered by Anonymous

To find the distance between the two towers :

According to the diagram , the base of the ∆ ABC will be the distance between the two towers.

In the ∆ ABC ,

Since we are provided with the height and we have to find the base , so we will use here tan θ, as we know that :

$\bf{\tan\:\theta = \dfrac{P}{B}}$

Using tan θ and substituting the values in it, we get :

$:\implies \bf{\tan60^{\circ} = \dfrac{45}{B}} \\ \\ \\$

$:\implies \bf{\sqrt{3} = \dfrac{45}{B}} \\ \\ \\$

By multiplying √3 to both the Numerator and Denominator , we get :

$:\implies \bf{B = \dfrac{45 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}} \\ \\ \\$

$:\implies \bf{B = \dfrac{45\sqrt{3}}{3}} \\ \\ \\$

$:\implies \bf{B = 15\sqrt{3}} \\ \\ \\$

$\boxed{\therefore \bf{B = 15\sqrt{3}\:m}} \\ \\ \\$

Hence , the base/distance between the two towers is 15√3 m.

To find the height of the second tower :

According to the diagram , the base of the ∆ DBC will be the height of the other towers.

In the ∆ DBC ,

Since we are provided with the base and we have to find the Height , so we will use here tan θ, as we know that :

$\bf{\tan\:\theta = \dfrac{P}{B}}$

Using tan θ and substituting the values in it, we get :

$:\implies \bf{\tan30^{\circ} = \dfrac{P}{15\sqrt{3}}} \\ \\ \\$

$:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{P}{15\sqrt{3}}} \\ \\ \\$

$:\implies \bf{\dfrac{15\sqrt{3}}{\sqrt{3}} = P} \\ \\ \\$

$:\implies \bf{15 = P} \\ \\ \\$

$\boxed{\therefore \bf{P = 15\:m}} \\ \\ \\$

Hence the height of the other tower is 15 m.

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