MCQ
Choose the correct answer from the given four options. If $\triangle\text{ABC}$ is right angled at $C,$ then the value of $\cos\text{(A+ B)}$ is :
- ✓$0$
- B$1$
- C$\frac{1}{2}$
- D$\frac{\sqrt{3}}{2}$
✓
Answer
Correct option: A.
$0$
We know that, in $\triangle\text{ABC,}$ sum of three angles $= 180^\circ $
$\text{i.e., }\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
But right angled at $C$.
i.e., $\angle\text{C}=90^\circ\ [$given$]$
$\angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\ \ \text{A}+\text{B}=90^\circ$
$[\because\angle\text{A}=\text{A}]$
$\therefore\ \ \cos(\text{A}+\text{B})=\cos90^\circ=0$
$\text{i.e., }\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
But right angled at $C$.
i.e., $\angle\text{C}=90^\circ\ [$given$]$
$\angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\ \ \text{A}+\text{B}=90^\circ$
$[\because\angle\text{A}=\text{A}]$
$\therefore\ \ \cos(\text{A}+\text{B})=\cos90^\circ=0$
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