Question

In $\Delta PQR$ <p=2 <Q and 2 < R=3 < Q, Calculate the angles of $\Delta PQR$.

• A. $<P={80}^0;<Q={40}^0;<R={60}^0$
• B. $<P={100}^0;<Q={40}^0;<R={60}^0$
• C. $<P={80}^0;<Q={60}^0;<R={50}^0$
• D. $<P={80}^0;<Q={40}^0;<R={30}^0$

Answer 1

The correct option is A $<P={80}^0;<Q={40}^0;<R={60}^0$

Given

$\angle P=2\angle Q$.......(1) and

$\angle R=\frac{3}{2}\angle Q$........(2)

Also sum of angles of a triangle is $180$ degree.

From equation (1) and (2):

$⟹\angle P+\angle Q+\angle R={180}^o$

$⟹2\angle Q+\left(3/2\right)\angle Q+\angle Q={180}^o$

$⟹\angle Q=\frac{{180}^o\times 2}{9}={40}^o$.

So $\angle P=2\left({40}^o\right)={80}^o$ and $\angle R=\left(3/2\right){20}^o={60}^o$