Properties of Triangles 1 Question 17
17. In a $\triangle A B C, A D$ is the altitude from $A$. Given $b>c, \angle C=23^{\circ}$ and $A D=\frac{a b c}{b^{2}-c^{2}}$, then $\angle B=\ldots$.
(1994, 2M)
Show Answer
Answer:
Correct Answer: 17. $113^{\circ}$
Solution:
- In $\triangle A D C, \frac{A D}{b}=\sin 23^{\circ}$
$$ \begin{array}{rlrl} \Rightarrow & A D & =b \sin 23^{\circ} \\ \text { But } & A D & =\frac{a b c}{b^{2}-c^{2}} \\ \Rightarrow & \frac{a b c}{b^{2}-c^{2}} & =b \sin 23^{\circ} \\ \Rightarrow & & \frac{a}{b^{2}-c^{2}} & =\frac{\sin 23^{\circ}}{c} \end{array} $$
[given]
Again, in $\triangle A B C$,
$$ \begin{aligned} & \frac{\sin A}{a} & =\frac{\sin 23^{\circ}}{c} \\ \Rightarrow & \frac{\sin A}{a} & =\frac{a}{b^{2}-c^{2}} \\ \Rightarrow & \sin A & =\frac{a^{2}}{b^{2}-c^{2}} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \quad \sin A=\frac{k^{2} \sin ^{2} A}{k^{2} \sin ^{2} B-k^{2} \sin ^{2} C} \\ & \Rightarrow \quad \sin A=\frac{\sin ^{2} A}{\sin ^{2} B-\sin ^{2} C} \\ & \Rightarrow \quad \sin A=\frac{\sin ^{2} A}{\sin (B+C) \sin (B-C)} \\ & \Rightarrow \quad \sin A=\frac{\sin ^{2} A}{\sin A \cdot \sin (B-C)} \\ & \Rightarrow \quad \sin (B-C)=1 \quad[\because \sin A \neq 0] \\ & \Rightarrow \quad \sin \left(B-23^{\circ}\right)=\sin 90^{\circ} \\ & \therefore \quad B-23^{\circ}=90^{\circ} \\ & \therefore \quad B=113^{\circ} \end{aligned} $$
