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The Height of a Tower is 100 M. When the Angle of Elevation of the Sun Changes from 30° to 45°, the Shadow of the Tower Becomes X Metres Less. the Value of X is - Mathematics

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Question

The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is 

Options

  • 100 m

  • \[100\sqrt{3} m\]

  • \[100\left( \sqrt{3} - 1 \right) m\]

  • \[\frac{100}{3}m\]

MCQ

Solution

The given situation can be represented as, 

Here, AB is the tower of height 100 meters.

When angle of elevation of sun changes from`∠D=30°` to `∠C=45°`, .`CD=x`

We assumed that  `BC=y`

Here we have to find the value of x

So we use trigonometric ratios.

In a triangle,`ABC` 

`⇒ tan C=(AB)/(BC)` 

`⇒ tan 45°=(AB)/(BC)` 

`⇒1=100/y`

`⇒y=100` 

Again in a triangle ABD,

`⇒ tan D=( AB)/(BC+CD)`

`⇒ tan 30°=100/(x+y)`

`⇒ 1/sqrt3=100/(x+y)` 

`⇒ 100sqrt3=x+y`

`⇒100sqrt3=x+100`                          `Put  x=100` 

`⇒x=100(sqrt3-1)`

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Chapter 12: Trigonometry - Exercise 12.3 [Page 42]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 12 Trigonometry
Exercise 12.3 | Q 12 | Page 42

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