Question

A gable roof (isosceles triangle-shaped) has a vertical height of 2.1 metres and the ceiling is 10.9 meters from one side to the other. Find the pitch (angle) of the roof.

Answer 1

Answer:

$\bold{21.07^\circ}$

Step-by-step explanation:

The given values can be mapped to an isosceles $\triangle ABC$.

Side AB = AC

Vertical height, AD = 2.1 m

The distance between one side to the other side of ceiling = 10.9 m

To find:

Pitch (Angle of the roof ) = ?

i.e. $\angle B$ or $\angle C$ = ? (because it is isosceles triangle, so both will be equal)

Solution:

As $\triangle ABC$ is isosceles, so vertical height will divide the side BC in two equal parts

i.e. $BD = DC = \frac{1}{2} BC$

$\therefore BD = \frac{10.9}{2} = 5.45 m$

In $\triangle ABD$, let us use tangent trigonometric property.

$tan\theta = \dfrac{Perpendicular}{Base}$

$tanB = \dfrac{AD}{BD}\\\Rightarrow tanB = \dfrac{2.1}{5.45}\\\Rightarrow tanB = 0.385\\\Rightarrow \angle B = tan^{-1}( 0.385)\\\Rightarrow \bold{\angle B = 21.07^\circ}$