Properties of Triangles 2 Question 10
10. Let $A, B, C$ be three angles such that $A=\frac{\pi}{4}$ and $\tan B, \tan C=p$. Find all positive values of $p$ such that $A, B, C$ are the angles of triangle.
(1997C, 5M)
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Solution:
- Since, $A+B+C=\pi$
$\Rightarrow \quad B+C=\pi-\pi / 4=3 \pi / 4$
$\therefore \quad 0<B, C<3 \pi / 4$
$[\because A=\pi / 4$, given $]$
Also, given $\quad \tan B \cdot \tan C=p$
$\Rightarrow \quad \frac{\sin B \cdot \sin C}{\cos B \cdot \cos C}=\frac{p}{1}$
$\Rightarrow \frac{\sin B \cdot \sin C+\cos B \cos C}{\sin B \cdot \sin C-\cos B \cdot \cos C}=\frac{p+1}{p-1}$
$\Rightarrow \quad \frac{\cos (B-C)}{\cos (B+C)}=\frac{1+p}{1-p}$
$$ \Rightarrow \quad \cos (B-C)=-\frac{(1+p)}{\sqrt{2}(1-p)} $$
$$ [\because B+C=3 \pi / 4] $$
Since, $B$ or $C$ can vary from 0 to $3 \pi / 4$
$$ \begin{array}{ll} \therefore & 0 \leq B-C<3 \pi / 4 \\ \Rightarrow & -\frac{1}{\sqrt{2}}<\cos (B-C) \leq 1 \end{array} $$
From Eqs. (ii) and (iii), $-\frac{1}{\sqrt{2}}<\frac{1+p}{\sqrt{2}(p-1)} \leq 1$
$$ \begin{aligned} & \Rightarrow \quad-\frac{1}{\sqrt{2}}<\frac{1+p}{\sqrt{2}(p-1)} \quad \text { and } \quad \frac{1+p}{\sqrt{2}(p-1)} \leq 1 \\ & \Rightarrow \frac{1+p}{p-1}+1 \geq 0 \text { and } \frac{1+p-\sqrt{2} p+\sqrt{2}}{\sqrt{2}(p-1)} \leq 0 \\ & \Rightarrow \quad \frac{2 p}{p-1} \geq 0 \text { and } \frac{(1-\sqrt{2}) p-\frac{1+\sqrt{2}}{1-\sqrt{2}}}{\sqrt{2}(p-1)} \leq 0 \\ & \Rightarrow \quad \frac{2 p}{p-1}>0 \quad \text { and } \quad \frac{\left(p-(\sqrt{2}+1)^{2}\right)}{(p-1)} \geq 0 \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \quad(p<0 \text { or } p>1) \\ & \text { and } \quad\left(p<1 \text { or } p>(\sqrt{2}+1)^{2}\right) \end{aligned} $$
On combining above expressions, we get
$$ \begin{array}{ll} & p<0 \text { or } p \geq(\sqrt{2}+1)^{2} \\ \text { i.e. } & p \in(-\infty, 0) \cup\left[(\sqrt{2}+1)^{2}, \infty\right) \\ \text { or } & p \in(-\infty, 0) \cup[3+2 \sqrt{2}, \infty) \end{array} $$
