(B) Let the height of the tower be $h = 60 \, m$ and the distance of the point from the base be $x$. In the right-angled triangle formed by the tower

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(B) Let the height of the tower be $h = 60 \, m$ and the distance of the point from the base be $x$. In the right-angled triangle formed by the tower and the ground,we have $\tan(60^{\circ}) = \frac{\text{height}}{\text{base}}$. Substituting the values,$\sqrt{3} = \frac{60}{x}$. Thus,$x = \frac{60}{\sqrt{3}} = \frac{60 \times \sqrt{3}}{3} = 20 \times \sqrt{3}$. Using $\sqrt{3} \approx 1.732$,we get $x = 20 \times 1.732 = 34.64 \, m$. Rounding to one decimal place,the distance is $34.6 \, m$.

Class 10 Mathematics Some Applications of Trigonometry Mix Examples - Some Applications of Trigonometry

Medium

The height of a tower is $60 \, m$. From a point on the ground,the angle of elevation of the top of the tower is found to be $60^{\circ}$. Find the distance of that point from the base of the tower. (in $m$)

A
$43.6$
B
$34.6$
C
$41.2$
D
$24.5$

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Some Applications of Trigonometry

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Mix Examples - Some Applications of Trigonometry

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The height of a tower is $60 \, m$. From a point on the ground,the angle of elevation of the top of the tower is found to be $60^{\circ}$. Find the distance of that point from the base of the tower. (in $m$)

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The angles of depression of two cars parked in the same direction of a $50\, m$ high tower are found to be $30^{\circ}$ and $60^{\circ}$. Find the distance between these two cars. (in $, m$)

Watching from a point $A$ on the ground $x \ m$ away from the base of the building. The height of the building is $y$ and the angle of elevation of the top of the building is found to be $25^\circ$. Then,$\ldots \ldots \ldots \ldots .$

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