Question

If $△ABC$ is a right angled triangle, right angled at C, then $\cos (A+B)=$

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (A)

$0$

Given that $△ABC$ is a right angled triangle, with $\angle C={90}^o$

To find out: The value of $\cos (A+B)$

In $△ABC$,

$\angle A+\angle B+\angle C={180}^o\left[\text{Angle sum property}\right]$

$\Rightarrow \angle A+\angle B+{90}^o={180}^o$

$\Rightarrow \angle A+\angle B={180}^o-{90}^o$

$\therefore \angle A+\angle B={90}^o$

Now, $\cos (A+B)=\cos {90}^o$

We know that, $\cos {90}^o=0$

$\therefore \cos (A+B)=0$

Hence, the required value of $\cos (A+B)$ for the given triangle is $0$.