Question:
If ΔABC is right angled at C, then the value of cos (A + B) is
(a) 0
(b) 1
(c) $\frac{1}{2}$
(d) $\frac{\sqrt{3}}{2}$
Solution:
(a) We know that, in $\triangle A B C$, sum of three angles $=180^{\circ}$
i.e., $\quad \angle A+\angle B+\angle C=180^{\circ}$
But right angled at $C$ i.e., $\angle C=90^{\circ}$ [given]
But right angled at $C$ i.e., $\angle C=90^{\circ}$
$\angle A+\angle B+90^{\circ}=180^{\circ}$
$\Rightarrow$ $A+B=90^{\circ}$ $[\because \angle A=A]$
$\therefore$ $\cos (A+B)=\cos 90^{\circ}=0$
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