Question

The angle of elevation of a ladder leaning against a wall is \[60{}^\circ \] and the foot of the ladder is 7.5 meters away from the wall. Find the length of the ladder.

Answer 1

Hint: First of all, assume that AB is the wall and BC is the distance between the foot of the ladder and wall, and AC is the length of the ladder. Now, in \[\Delta ABC\] we have, \[\angle ACB=60{}^\circ \] , Base = BC = 7.5 meters, and Hypotenuse = AC. We know the formula for the cosine of an angle, \[\cos \theta =\dfrac{Base}{Hypotenuse}\] . Now, use this cosine formula and use \[\cos 60{}^\circ =\dfrac{1}{2}\] to simplify it. Then, solve it further and get the value of AC.

Complete step by step answer:
According to the question, it is given that
The angle of elevation of a ladder leaning against a wall = \[60{}^\circ \] …………………………………………(1)
The distance between the foot of the ladder and wall = 7.5 meters ………………………………………..(2)
Let us assume that AB is the wall, BC is the distance between the foot of the ladder and wall, and AC is the length of the ladder.

In \[\Delta ABC\] we have,
\[\angle ACB=60{}^\circ \] ………………………………………………..(3)
Base = BC = 7.5 meters ………………………………………………(4)
Hypotenuse = AC ……………………………………………………(5)
We know the formula for the cosine of an angle, \[\cos \theta =\dfrac{Base}{Hypotenuse}\] ……………………………………(6)
Now, from equation (3), equation (4), equation (5), and equation (6), we get
\[\cos 60{}^\circ =\dfrac{BC}{AC}\]
\[\Rightarrow \cos 60{}^\circ =\dfrac{7.5}{AC}\] …………………………………….(7)
We also know that \[\cos 60{}^\circ =\dfrac{1}{2}\] ………………………………………………(8)
Now, using equation (8) and on substituting \[\cos 60{}^\circ \] by \[\dfrac{1}{2}\] in equation (7), we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}=\dfrac{7.5\,meters}{AC} \\
 & \Rightarrow AC=7.5\times 2\,meters\, \\
\end{align}\]
\[\Rightarrow AC=15\,meters\] ……………………………………..(9)
Since AC is the length of the ladder and from equation (9), we have the length of side AC.

Therefore, the length of the ladder is 15 meters.

Note: We can also solve this question by another method.

 In the \[\Delta ABC\] we have,
\[\angle ACB=60{}^\circ \] ………………………………………………..(1)
\[\angle ABC=90{}^\circ \] ……………………………………………….(2)
BC = 7.5 meters ………………………………………………(3)
We know the property that the sum of all angles of a triangle is \[180{}^\circ \] .
Using this property, \[\angle ACB+\angle ABC+\angle CAB=180{}^\circ \] ………………………………………..(4)
Now, from equation (1), equation (2), and equation (4), we get
 \[\begin{align}
  & \Rightarrow 60{}^\circ +90{}^\circ +\angle CAB=180{}^\circ \\
 & \Rightarrow \angle CAB=180{}^\circ -150{}^\circ \\
\end{align}\]
\[\Rightarrow \angle CAB=30{}^\circ \] …………………………………………..(5)
Now, in \[\Delta ABC\] for \[\angle CAB\] , we have
Perpendicular = BC = 7.5 meters ……………………………………………(6)
 Hypotenuse = AC ………………………………………….(7)
We know the formula for the sine of an angle, \[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}\] ………………………………………(8)
Now, from equation (5), equation (6), equation (7), and equation (8), we get
\[\sin 30{}^\circ =\dfrac{7.5\,meters}{AC}\] ……………………………………………(9)
We also know that \[\sin 30{}^\circ =\dfrac{1}{2}\] ……………………………………………….(10)
Now, from equation (9) and equation (10), we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}=\dfrac{7.5\,meters}{AC} \\
 & \Rightarrow AC=2\times 7.5\,meters \\
 & \Rightarrow AC=15\,meters \\
\end{align}\]
Therefore, the length of the ladder is 15 meters.