Question

the angle of elevation of the top of the tower from a point a on the ground which is 30 metre away from the foot of the tower is 30° find the height of the tower​

Answer 1

Given :

• The tower is making an angle of elevation of 30° to the ground.
• The angle is made 30 m away from the foot of the tower.

To Find :

The height of the tower.

Solution :

Analysis :

Here the concept of heights and distances is used. We have to use the concept of angle of elevation along with trigonometric ratios to find the height of the tower.

Explanation :

Let BC be the h metres and A be the point on the ground 30 m away from the foot of the tower.

Then, Angle if elevation = ∠CAB = 30°.

We have this concept that,

$\boxed{\bf\dfrac{Required\ Side}{Known\ Side}=a\ certain\ t-ratio\ of\ a\ known\ angle}$

So,

In ∆ABC,

The base is given and we have to find the height.

We know that,

$\boxed{\bf\tan\theta=\dfrac{Height}{Base}}$

• We know that tan30° = 1/√3.

In ∆ABC,

• ∠ABC = 90°

From ∆ABC we get,

$\\ \implies\bf\tan\theta=\dfrac{Height}{Base}$

where,

• tanθ = 30° = 1/√3
• Height = h m
• Base = 30 m

Using the required formula and substituting the required values,

$\\ \implies\sf\tan30^{\circ}=\dfrac{h}{30}$

Putting tan30° = 1/√3,

$\\ \implies\sf\dfrac{1}{\sqrt{3}}=\dfrac{h}{30}$

By cross multiplying,

$\\ \implies\sf1\times30=\sqrt{3}\times h$

$\\ \implies\sf30=\sqrt{3}h$

$\\ \implies\sf\dfrac{30}{\sqrt{3}}=h$

Rationalizing the denominator,

$\\ \implies\sf\dfrac{30\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=h$

$\\ \implies\sf\dfrac{30\sqrt{3}}{3}=h$

$\\ \implies\sf\dfrac{\cancel{30\sqrt{3}}\ \ \ ^{10\sqrt{3}}}{\not{3}}=h$

$\\ \implies\sf10\sqrt{3}=h$

$\\ \therefore\boxed{\bf Height=10\sqrt{3}.}$

The height of the tower is 10√3.

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Explore More :

The trigonometric table :

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

Trigonometric Ratios :

$\\ \bullet\ \sf\sin\theta=\dfrac{Height}{Hypotenuse}$

$\\ \bullet\ \sf\cos\theta=\dfrac{Base}{Hypotenuse}$

$\\ \bullet\ \sf\tan\theta=\dfrac{Height}{Base}$

$\\ \bullet\ \sf\cot\theta=\dfrac{Base}{Height}$

$\\ \bullet\ \sf\sec\theta=\dfrac{Hypotenuse}{Base}$

$\\ \bullet\ \sf\cosec\theta=\dfrac{Hypotenuse}{Height}$