Question

In ΔPQR, if 3∠P = 4∠Q = 6∠R, calculate the angles of the triangle.

Answer 1

$\textbf{\huge{ANSWER:}}$

$\tt{Given:}$

3 (Angle P) = 4 (Angle Q) = 6 (Angle R) ...(1)

Now, we know that the $\textbf{sum of all the angles is = 180°}$ (The angle sum property )

Thus;
Angles ( P + Q + R ) = 180°

Multiply the equation by 3:-

=》3 [Angles ( P + Q + R )] = 180° × 3

=》 Angles ( 3P + 3Q + 3R ) = 540°

According to (1) :-

=》 Angles ( 4Q + 3Q + 3R ) = 540°

=》 Angles ( 7Q + 3R ) = 540°

Multiply the equation by 2:-

=》 2 [Angles ( 7Q + 3R )] = 540° × 2

=》 Angles ( 14Q + 6R ) = 1080°

According to (1) :-

=》 Angles ( 14Q + 4Q ) = 1080°

=》 Angles ( 18Q ) = 1080°

=》 Angle Q = $\frac{1080°}{18}$

=》 $\textbf{ Angle Q = 60°}$

Now, Given is that:

3 (Angle P) = 4 (Angle Q) = 6 (Angle R)

=》 3 (Angle P) = 4 ( 60° ) = 6 (Angle R)

=》 3 (Angle P) = 240° = 6 (Angle R)

Taking both of them separately :-

=》 3P = 240°

=》 P = $\frac{240°}{3}$

=》 $\textbf{P = 80°}$

Next one :-

=》 6R = 240°

=》 R = $\frac{240°}{6}$

=》 $\textbf{R = 40°}$