Question

In a triangle $ABC,AD$ is the altitude from $A$. Given $b>c$, angle $C={23}^{\circ }$ and $AD=\frac{abc}{b^2-c^2}$ then angle $B=$

• A. ${157}^{\circ }$
• B. ${113}^{\circ }$
• C. ${147}^{\circ }$
• D. None

Answer 1

The correct option is B ${113}^{\circ }$

In $△ADC,$

$\Rightarrow$ $\sin {23}^o=\frac{AD}{b}$

$\Rightarrow$ $AD=b\sin {23}^o$

But $AD=\frac{abc}{b^2-c^2}$ [ Given ]

$\Rightarrow$ $\frac{abc}{b^2-c^2}=b\sin {23}^o$

$\Rightarrow$ $\frac{a}{b^2-c^2}=\frac{\sin {23}^o}{c}$ ----- ( 1 )

Again, In $△ABC,$

$\frac{\sin A}{a}=\frac{\sin {23}^o}{c}$

From equatin ( 1 ),

$\Rightarrow$ $\frac{\sin A}{a}=\frac{a}{b^2-c^2}$

$\Rightarrow$ $\sin A=\frac{a^2}{b^2-c^2}$

$\Rightarrow$ $\sin A=\frac{k^2{\sin }^{2}A}{k^2{\sin }^{2}B-k^2{\sin }^{2}C}$

$\Rightarrow$ $\sin A=\frac{{\sin }^{2}A}{{\sin }^{2}B-{\sin }^{2}C}$

$\Rightarrow$ $\sin A=\frac{{\sin }^{2}A}{\sin (B+C)\sin (B-C)}$

$\Rightarrow$ $\sin A=\frac{{\sin }^{2}A}{\sin A.\sin (B-C)}$

$\Rightarrow$ $\sin (B-C)=1$ [ Since, $\sin A\ne 0$ ]

$\Rightarrow$ $\sin (B-{23}^o)=\sin {90}^o$

$\Rightarrow$ $B-{23}^o={90}^o$

$\Rightarrow$ $B={113}^o$