Question

A ladder placed against a wall such that it reaches the top of the wall of height 6m and the ladder is inclined at an angle of 60 degree. Find how far the ladder is from the foot of the wall

Answer 1

Step-by-step explanation:

Given:

In figure, AB represent the wall such that AB=6m, a ladder is placed against the wall at an angle of 60°,angle ACB=60°

To find:

BC

Solution:

Angle ACB=60°........[given]

By applying formula,

$\tan(60) = \frac{opposite \: side}{adjcent \: side}$

$\frac{ \sqrt{3} }{1} = \frac{ab}{bc}$

√3 = 6

1 BC

BC x √3 = 6

BC = 6.

√3

Rationalize the denominator

BC = 6. x √3

√3 √3

BC= 6 x √3.............. [√3 x √3=3]

3

BC= 2√3

If √3 value is given 1.732

Then,

BC = 2 x 1.732

BC = 3.464

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Answer 2

Concept:

The angle created by a given line and the x-axis is known as the angle of inclination

Given:

The height of the wall is $6m$ and the angle of inclination is $60^{\circ}$

To find:

Distance between the foot of the ladder and the wall.

Solution:

The angle of Inclination is $60^{\circ}$.

Therefore, finding tan$60^{\circ}$.

$\tan (60)=\frac{\text { Perpendicular }}{\text { Base}}\\{\frac{\sqrt{3}}{1}}=\frac{AB}{BC}\\\frac{6}{\sqrt{3} } =BC \\\\BC=2\sqrt{3}$

Therefore, the distance between the foot of the ladder and the wall is $2 \sqrt{3}m$.

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