Question

Question 8
If $\Delta ABC$ is right angled at C, then the value of cos(A+B) is

(A) 0
(B) 1
(C) $\frac{1}{2}$
(D) $\frac{\sqrt{3}}{2}$

Answer 1

The answer is (A).

We know that, in $\Delta ABC$, sum of three angles $={180}^{\circ }$

i.e., $\angle A+\angle B+\angle C={180}^{\circ }$

But right angled at C i..e, $\angle C={90}^{\circ }$ [given]

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow A+B={90}^{\circ }$

$\therefore cos(A+B)=cos{90}^{\circ }\lfloor \because \angle A=A\rfloor$