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

• A

  $\frac{\sqrt{5}+1}{2}$ and $\frac{\sqrt{5}-1}{2}$
• Correct answer
  B

  $\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$
• C

  $\sqrt{5}$ and $\frac{1}{\sqrt{5}}$
• D

  $\frac{\sqrt{5}}{2}$ and $\frac{2}{\sqrt{5}}$

Solution

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 $^{2}$ $\begin{array}{lr}\text { Now, } & \begin{array}{r}A C^{2}=A B^{2}+B C^{2} \\ \left(a r^{2}\right)^{2}=a^{2}+(a \cdot r)^{2}\end{array} \\ \Rightarrow \quad a^{2} r^{4}=a^{2}+a^{2} r^{2} \\ \Rightarrow \quad r^{4}-r^{2}-1=0 \\ \Rightarrow \quad r^{2}=\frac{1+\sqrt{5}}{2} \\ \Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}\end{array}$ $\therefore \quad \tan \alpha=\frac{B C}{A B}=\frac{a r}{a}$ $\Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}$ $\begin{aligned} \text { and } \tan \beta &=\frac{A B}{B C}=\frac{a}{a r}=\frac{1}{r} \\ &=\frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}}=\sqrt{\frac{\sqrt{5}-1}{2}} \end{aligned}$

Hence, the correct answer is option: 'B'.