If $ \triangle \mathrm{ABC} $ is right angled at $ \mathrm{C} $, then the value of $ \cos (\mathrm{A}+\mathrm{B}) $ is
(A) 0
(B) 1
(C) $ \frac{1}{2} $
(D) $ \frac{\sqrt{3}}{2} $
Given:
\( \triangle \mathrm{ABC} \) is right angled at \( \mathrm{C} \)
To do:
We have to find the value of \( \cos (\mathrm{A}+\mathrm{B}) \).
Solution:
\( \triangle \mathrm{ABC} \) is right angled at \( \mathrm{C} \)
This implies,
$\angle C=90^{\circ}$
We know that,
In $\triangle A B C$,
Sum of the three angles $=180^{\circ}$
$\angle A+\angle B+\angle C=180^{\circ}$
$\angle A+\angle B+90^{\circ} =180^{\circ}$
$A+B=180^{\circ}-90^{\circ}$
$A+B=90^{\circ}$
Therefore,
$\cos (A+B)=\cos 90^{\circ}=0$
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