Question

Assertion :In a triangle ABC
$\sin A+\sin B+\sin C=1+\frac{\sqrt{3}}{2}$, then $\sin A=\sin B$ Reason: In triangle ABC, ${(a+b)}^2=c^2+ab$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion false but Reason is true

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

${(a+b)}^2=c^2+ab$

$\Rightarrow a^2+b^2-c^2=-ab$

$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}$

$\Rightarrow \cos C=-\frac{1}{2}$

$\Rightarrow C={120}^{\circ }$
Now we have $\sin A+\sin B+\sin C=1+\frac{\sqrt{3}}{2}$

$\Rightarrow \sin A+\sin B=1$ ($\because sin{120}^0=\frac{\sqrt{3}}{2}$)

$\Rightarrow 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=1$

$\Rightarrow 2\cos \frac{C}{2}\cos \left(\frac{A-B}{2}\right)=1$

$\Rightarrow \cos \left(\frac{A-B}{2}\right)=\cos 0^{\circ }$

$\Rightarrow A=B$

$\Rightarrow \sin A=\sin B$