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} $

AcademicMathematicsNCERTClass 10

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,

$\cos (A+B)=\cos 90^{\circ}=0$

raja
Updated on 10-Oct-2022 13:28:52

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