Search any question & find its solution
Question:
Answered & Verified by Expert
The three sides of a right angled triangle are in GP (geometric progression). If the two acute angles be $\alpha$ and $\beta$, then tan $\alpha$ and tan $\beta$ are
Options:
Solution:
1348 Upvotes
Verified Answer
The correct answer is:
$\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$
Let $\Delta A B C$ be a right angled triangle at $B$. Let $\angle A$ and $\angle C$ be $\alpha$ and $\beta$
Since, sides are in GP so sides are $a,$ ar, ar $^{2}$
$\begin{array}{lr}\text { Now, } & \begin{array}{r}A C^{2}=A B^{2}+B C^{2} \\ \left(a r^{2}\right)^{2}=a^{2}+(a \cdot r)^{2}\end{array} \\ \Rightarrow \quad a^{2} r^{4}=a^{2}+a^{2} r^{2} \\ \Rightarrow \quad r^{4}-r^{2}-1=0 \\ \Rightarrow \quad r^{2}=\frac{1+\sqrt{5}}{2} \\ \Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}\end{array}$
$\therefore \quad \tan \alpha=\frac{B C}{A B}=\frac{a r}{a}$
$\Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}$
$\begin{aligned} \text { and } \tan \beta &=\frac{A B}{B C}=\frac{a}{a r}=\frac{1}{r} \\ &=\frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}}=\sqrt{\frac{\sqrt{5}-1}{2}} \end{aligned}$
Since, sides are in GP so sides are $a,$ ar, ar $^{2}$
$\begin{array}{lr}\text { Now, } & \begin{array}{r}A C^{2}=A B^{2}+B C^{2} \\ \left(a r^{2}\right)^{2}=a^{2}+(a \cdot r)^{2}\end{array} \\ \Rightarrow \quad a^{2} r^{4}=a^{2}+a^{2} r^{2} \\ \Rightarrow \quad r^{4}-r^{2}-1=0 \\ \Rightarrow \quad r^{2}=\frac{1+\sqrt{5}}{2} \\ \Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}\end{array}$
$\therefore \quad \tan \alpha=\frac{B C}{A B}=\frac{a r}{a}$
$\Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}$
$\begin{aligned} \text { and } \tan \beta &=\frac{A B}{B C}=\frac{a}{a r}=\frac{1}{r} \\ &=\frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}}=\sqrt{\frac{\sqrt{5}-1}{2}} \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.