MCQ
In a $\triangle A B C$, right angled at $B$, the value of $\sin (A+C)$ is
- A$0$
- ✓1
- C$\frac{1}{2}$
- D$\frac{\sqrt{3}}{2}$
✓
Answer
Correct option: B.
1
(B)1
In $\triangle A B C$, it is given that $\angle B=90^{\circ}$
$
\therefore \quad A+B+C=180^{\circ} \Rightarrow A+90^{\circ}+C=180^{\circ} \Rightarrow A+C=90^{\circ} \Rightarrow \sin (A+C)=\sin 90^{\circ}=1
$
In $\triangle A B C$, it is given that $\angle B=90^{\circ}$
$
\therefore \quad A+B+C=180^{\circ} \Rightarrow A+90^{\circ}+C=180^{\circ} \Rightarrow A+C=90^{\circ} \Rightarrow \sin (A+C)=\sin 90^{\circ}=1
$
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