Question

If $△PQR$ is right angled at R then the value of $\cos (P+Q)$ is

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A 0

$△$PQR is right angled at R then $\angle R={90}^{\circ }$

$cos(P+Q)=cos(180-R)$ (Since sum of the angle in a triangle is 180)

$\Rightarrow cos(P+Q)=cos(180-90)$

$\Rightarrow cos(P+Q)=cos{90}^{\circ }$

$\because cos{90}^{\circ }=0$

$\therefore cos(P+Q)=0$