Question

A ladder leaning against a vertical wall makes an angle of ${60}^o$ with the ground. The foot of the ladder is $3.5$ m away from the wall. Find the length of the ladder.

Answer 1

Let $AB$ be the length of the ladder and $CA$ be the wall.

The ladder makes an angle of ${60}^{\circ }$ with the horizontal.

$\therefore △ABC$ is ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ ,right triangle

We have $\tan {60}^{\circ }=\frac{AC}{BC}$

$\Rightarrow AC=3.5\times \tan {60}^{\circ }$

$\Rightarrow AC=3.5\times \sqrt{3}$m

From Pythagoras theorem, we have

${AB}^2={BC}^2+{CA}^2$

$\Rightarrow {AB}^2={\left(3.5\right)}^2+{(3.5\times \sqrt{3})}^2$

$\Rightarrow {AB}^2=12.25+36.75$

$\Rightarrow {AB}^2=49$

$\therefore AB=7m$

Hence,length of the ladder$=7m$