Question

If in a right angled triangle ABC, $4\sin A\cos B-1=$ 0 and $\tan A$ is real, then

• A. angles are in A.P.
• B. the triangle is isosceles
• C. $\sin A+\sin B+\sin C=\frac{3+\sqrt{3}}{2}$
• D. $\frac{a}{1}=\frac{b}{\sqrt{3}}=\frac{c}{2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A angles are in A.P.
C $\frac{a}{1}=\frac{b}{\sqrt{3}}=\frac{c}{2}$
D $\sin A+\sin B+\sin C=\frac{3+\sqrt{3}}{2}$

$4\sin A\cos B=1$
Or
$2\sin A.\cos B=\frac{1}{2}$
Or
$\sin (A+B)+\sin (A-B)=\frac{1}{2}$
Or
$\sin (\pi -C)+\sin (A-B)=\frac{1}{2}$
Or
$\sin \left(C\right)+\sin (A-B)=\frac{1}{2}$
Or
$\sin {90}^0+\sin (A-B)=\frac{1}{2}$
Or
$\sin (A-B)=-\frac{1}{2}$
Or
$A-B=-{30}^0$.
And $A+B={90}^0$
Hence
$A={30}^0$ and $B={60}^0$.
Hence the angles are ${30}^0,{60}^0,{90}^0$.