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  • Let in a right angled triangle, the smallest angle be . If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then <img alt="\sin \theta" src="/latex-image/?%5Csin

Let in a right angled triangle, the smallest angle be \theta. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then \sin \theta is equal to :
 
Option: 1 \frac{\sqrt{5}+1}{4}
Option: 2 \frac{\sqrt{5}-1}{2}
Option: 3 \frac{\sqrt{2}-1}{2}
 
Option: 4 \frac{\sqrt{5}-1}{4}

Answers (1)

best_answer

Here

\quad b<a<\sqrt{a^{2}+b^{2}}

Theor reciprocals:

\frac{1}{\sqrt{a^{2}+b^{2}}}<\frac{1}{a}<\frac{1}{b}.

These also form right angled triangle, so

\begin{aligned} \therefore & \frac{1}{a^{2}+b^{2}}+\frac{1}{a^{2}}=\frac{1}{b^{2}} \\ \end{aligned}

\begin{aligned} \Rightarrow & \frac{a^{2}+a^{2}+b^{2}}{a^{2}\left(a^{2}+b^{2}\right)}=\frac{1}{b^{2}} \\ \end{aligned}

\begin{aligned} \Rightarrow & 2 a^{2} b^{2}+b^{4}=a^{4}+a^{2} b^{2} \\ \end{aligned}

\begin{aligned} \Rightarrow & b^{4}+a^{2} b^{2}-a^{4}=0 \\ \end{aligned}

\begin{aligned} \Rightarrow &\left(\frac{b}{a}\right)^{4}+\left(\frac{b}{a}\right)^{2}-1=0 \quad\left(\text { dividing by } a^{4}\right) \\ \end{aligned}

\begin{aligned} \Rightarrow &\left(\frac{b}{a}\right)^{2}=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{\sqrt{5}-1}{2} \text { (negative rejected) } \end{aligned}

From first figure \frac{b}{a}=\tan \Theta

\begin{aligned} &\Rightarrow \tan ^{2} \theta=\frac{\sqrt{5}-1}{2} \\ &\Rightarrow \sec ^{2} \theta=\frac{\sqrt{5}+1}{2} \Rightarrow \cos ^{2} \theta=\frac{2}{\sqrt{5}+1} \\ &\Rightarrow \sin ^{2} \theta=\frac{\sqrt{5}-1}{\sqrt{5}+1} \Rightarrow \sin \theta=\frac{\sqrt{5}-1}{2} \end{aligned}

Hence option (2) is correct.

Posted by

Kuldeep Maurya

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