Question

A 35 M long ladder just reach a window of a building. If the ladder makes an angle {theta} with the level of ground such that
$10 \: theta= \frac{1}{2 \sqrt{6} }$
then find
1)-height of the window above the ground level
2)-horizontal distance of the foot of the ladder from the building

Answer 1

i think that 10 theta is actually tan theta

Answer 2

Concept:

The Pythagorean theorem, sometimes known as Pythagoras' theorem, is a fundamental relationship between the three sides of a right triangle in Euclidean geometry. The size of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides, according to this rule.

Given:

A ladder length is $35m$ making an angle $tan\theta=\frac{1}{2\sqrt{6}}$

To Find:

1. The height of the window above the ground level.

2. The horizontal distance of the foot of the ladder from the building.

Solution:

The angle given is$tan\theta=\frac{1}{2\sqrt{6}}$.

$\tan \theta=\frac{1}{2 \sqrt{6}}=\frac{A C}{B C}\\\\ B C=A C \times 2 \sqrt{6} \\\\ (B C)^{2}=24(A C)^{2}$

Use Pythagoras theorem

$A B^{2}=A C^{2}+B C^{2}$

Put values in it.

$35 \times 35=25(A C)^{2}\\ (AC)^{2}=\frac{35 \times 35}{25} \\\\ (A C)^{2}=49 \\AC=7$

Therefore the height of the window above the ground level is $7m$.

Find the value of $BC$ to calculate the horizontal distance of the foot of the ladder.

Put the value of $AC$ in $B C=A C \times 2 \sqrt{6}$.

Therefore $B C=14 \sqrt{6}$

The horizontal distance of the foot of the ladder from the building is $14 \sqrt{6}m$                                                                                                                                          

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