Question

in∆PQR, LP=2LQ and 2LR=3LQ.calculate the angles of ∆PQR​

Answer 1

GIVEN :  

$\sf \: P\: = \: 2\: \angle \: Q$

$\sf \: 2 \: \angle\:R \: =\:3 \:\angle \:Q$

What to Calculate ?

$\sf \: We\:need \: to \: calculate \: the \: angles\:of \: \triangle PQR$

So, Let's Start :

$\sf \: P\: = \: 2\: \angle\: Q$

$\sf \: 2 \: \angle \:R \: =\:3 \:\angle \:Q$

$\sf \: \angle \:R \: = \: \dfrac{3}{2} \:\angle \: \:Q$

$\sf \:\angle \: P\: + \: \angle \: Q\: + \:\angle\:R = 180^{ \circ}$

$\sf \: 2 \: \angle\: Q \: + \: \angle\: Q\: \: + \: \dfrac{3}{2} \: \angle \: Q \: = 180^{ \circ}$

$\dfrac{\sf \: 4 \: \angle\:Q \: + \: 2 \: \angle \:Q\: + \: 3 \:\angle \: Q \:}{2} = \: 180^{ \circ}$

$\sf \: 9\: \angle \:Q \: = \: 360^{ \circ}$

$\sf \: \angle\:Q \: = \: \dfrac{360}{9} \: = \: 40^{ \circ}$

$\sf \: \angle\: P \: = \: 2\: \times \:40 \: \longrightarrow \:80^{ \circ}$

$\sf \: \angle\: R \: = \: \dfrac{3}{2} \times 40 \: \longrightarrow \:60^{ \circ}$