Question

$△ABC$ is a right-angled triangle with sides $a,b,c$ and smallest angle $\theta$. If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are also the sides of the right-angle triangle then find $\sin \theta$.

• A. $\sqrt{\frac{3-\sqrt{5}}{2}}$
• B. $\frac{3-\sqrt{5}}{2}$
• C. $\sqrt{\frac{3+\sqrt{5}}{2}}$
• D. $\frac{3+\sqrt{5}}{2}$

Answer 1

The correct option is A

$\sqrt{\frac{3-\sqrt{5}}{2}}$

Explanation for the correct option.

Step 1. Find the value of $\sin \theta$ in terms of $a,b,c$.

Let in the right angles triangle $a>b>c$ and as $\theta$ is the smallest angle, so the triangle can be drawn as:

Now, the value of $\sin \theta$ is given as: $\sin \theta =\frac{c}{a}$.

Also for the triangle $a$ is the hypotenuse, so using Pythagoras theorem $a^2=b^2+c^2$.

Step 2. Use the Pythagoras theorem for the other triangle.

As $a>b>c$, so$\frac{1}{c}>\frac{1}{b}>\frac{1}{a}$ and as $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ form a right angles triangle so side $\frac{1}{c}$ is the hypotenuse and the pythagoras theorem can be used as: $\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

Now it can be simplified as:

$$\begin{gathered} \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2} \\ \Rightarrow 1=\frac{c^2}{a^2}+\frac{c^2}{b^2} \\ \Rightarrow 1={\left(\frac{c}{a}\right)}^2+\frac{c^2}{a^2-c^2}\left[a^2=b^2+c^2\right] \\ \Rightarrow 1={\left(\frac{c}{a}\right)}^2+\frac{1}{{\left({\displaystyle \frac{a}{c}}\right)}^2-1} \\ \Rightarrow 1={\left(\sin \theta \right)}^2+\frac{1}{{\left(\csc \theta \right)}^2-1}\left[\sin \theta =\frac{c}{a},\csc \theta =\frac{1}{\sin \theta }\right] \\ \Rightarrow 1=\frac{1-{\sin }^2\theta +1}{{\csc }^2\theta -1} \\ \Rightarrow {\csc }^2\theta -1=2-{\sin }^2\theta \\ \Rightarrow {\sin }^2\theta +{\csc }^2\theta -3=0 \end{gathered}$$

Step 3. Form a quadratic equation in ${\sin }^2\theta$ and solve for $\sin \theta$.

In the equation ${\sin }^2\theta +{\csc }^2\theta -3=0$ use $\csc \theta =\frac{1}{\sin \theta }$ to form a quadratic equation in ${\sin }^2\theta$.

$$\begin{gathered} {\sin }^2\theta +{\csc }^2\theta -3=0 \\ \Rightarrow {\sin }^2\theta +\frac{1}{{\sin }^2\theta }-3=0 \\ \Rightarrow {\left({\sin }^2\theta \right)}^2-3{\sin }^2\theta +1=0 \end{gathered}$$

Now, the solution for the quadratic equation in ${\sin }^2\theta$ is given as:

$\begin{array}{rcl}{\sin }^2\theta & =& \frac{-(-3)\pm \sqrt{{(-3)}^2-4\times 1\times 1}}{2\times 1}\\ & =& \frac{3\pm \sqrt{9-4}}{2}\\ & =& \frac{3\pm \sqrt{5}}{2}\end{array}$

The value of ${\sin }^2\theta$ cannot be greater than $1$ and so $\begin{array}{rcl}{\sin }^2\theta & =& \frac{3+\sqrt{5}}{2}\end{array}$is rejected.

Thus $\begin{array}{rcl}{\sin }^2\theta & =& \frac{3-\sqrt{5}}{2}\end{array}$ and $\begin{array}{rcl}\sin \theta & =& \sqrt{\frac{3-\sqrt{5}}{2}}\end{array}$.

So the value of $\sin \theta$ is $\sqrt{\frac{3-\sqrt{5}}{2}}$.

Hence, the correct option is A.