Question

A ladder leaning against a wall makes an angle of $${60}^{\circ }$$ with the horizontal. If the fact of the ladder is 2.5 m away from the wall, find the length of the ladder.

Answer 1

- Hint:- Draw figure as per mentioned in the question. Consider the triangle formed. Take the cosine function and find the length of the ladder.

Complete step-by-step answer: -

From the figure you can understand that BC is the wall in which the ladder AC is leaning against.
It is said that the ladder AC makes $${60}^{\circ }$$ angle with the horizontal, i.e. the ladder is leaning against the wall at an angle of $${60}^{\circ }$$.
Now the distance between the foot of the ladder and the wall is 2.5 m. Thus from the figure, you can say that, AB = 2.5 m.
Now the wall is perpendicular to the ground. So BC is perpendicular to AB, i.e. $$BC\mathrm{\perp }AB$$ and $$\mathrm{\angle }ABC={90}^{\circ }$$.
Now let us consider $$\mathrm{\Delta }ABC$$, from the figure.
$$\begin{array}{rl}& \cos {60}^{\circ }={\displaystyle \frac{adjacentangle}{hypotenuse}}={\displaystyle \frac{AB}{AC}}\\ & \cos {60}^{\circ }={\displaystyle \frac{2.5}{AC}}\\ & \Rightarrow AC={\displaystyle \frac{2.5}{\cos {60}^{\circ }}}\end{array}$$
From trigonometry, we know that $$\cos {60}^{\circ }={}^1/{}_2$$.
$$\therefore AC={\displaystyle \frac{2.5}{{}^1/{}_2}}=2.5\times 2=5$$ m.
Thus we got AC = 5 cm.
So the length of the ladder is 5 cm.

Note:
We can find the length of the wall BC, by using Pythagoras theorem.
$$\begin{array}{rl}& A{B}^2+B{C}^2=A{C}^2\\ & B{C}^2=A{C}^2-A{B}^2\\ & \Rightarrow BC=\sqrt{A{C}^2-A{B}^2}=\sqrt{5^2-{\left(2.5\right)}^2}=\sqrt{25-6.25}=\sqrt{18.75}\\ & =4.33m\end{array}$$
Thus, the height of the wall is 4.33 m.