Question

In \[\vartriangle PQR\], $\dfrac{{PQ}}{1} = \dfrac{{PR}}{2} = \dfrac{{QR}}{{\sqrt 3 }}$, then $m\angle R$ is:
A. ${90^ \circ }$
B. ${60^ \circ }$
C. ${45^ \circ }$
D. ${30^ \circ }$

Answer 1

Hint: We will first use the ratios of sides given and write every side in terms of one only and then we will notice that it follows Pythagorean and we will get angle Q to be right angled. Then we will use trigonometric ratio to get the required angle.

Complete step-by-step answer:
We are already given that $\dfrac{{PQ}}{1} = \dfrac{{PR}}{2} = \dfrac{{QR}}{{\sqrt 3 }}$.
We will now write every side in terms of PQ.
Since, we have:- $\dfrac{{PQ}}{1} = \dfrac{{PR}}{2}$.
Cross multiplying it to get the result as follows:
$PR = 2PQ$ ……….(1)
Now, we also have:- $\dfrac{{PQ}}{1} = \dfrac{{QR}}{{\sqrt 3 }}$
Cross multiplying it to get the result as follows:
$QR = \sqrt 3 PQ$ ……….(2)
Squaring the equation (1), we will have with us the following result:-
$ \Rightarrow P{R^2} = {(2PQ)^2}$
$ \Rightarrow P{R^2} = 4P{Q^2}$ …………(3)
Squaring the equation (2), we will have with us the following result:-
$ \Rightarrow Q{R^2} = {(\sqrt 3 PQ)^2}$
$ \Rightarrow Q{R^2} = 3P{Q^2}$ ………..(4)
Now, if we add $P{Q^2}$ to LHS and RHS in the equation (4), we will get:-
$ \Rightarrow P{Q^2} + Q{R^2} = 3P{Q^2} + P{Q^2}$
Simplifying it to get:-
$ \Rightarrow P{Q^2} + Q{R^2} = 4P{Q^2}$
Now using (3), we can write it as:-
$ \Rightarrow P{Q^2} + Q{R^2} = P{R^2}$
So, we clearly see that the \[\vartriangle PQR\] follows the Pythagorean Theorem which states that:-
In a right angled triangle the square of the long side is equal to the sum of the squares of the other two sides. The long side is called the hypotenuse.
So, we now know that \[\vartriangle PQR\] is a right angled triangle.
Now, looking at this expression $P{Q^2} + Q{R^2} = P{R^2}$ we clearly see that PR is acting like a hypotenuse and rest are base and perpendicular. Now, let us draw it to get a clearer picture.
               

We now just need to find the measure of the angle R.
We know that $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
Therefore, $\tan R = \dfrac{{PQ}}{{QR}}$
Now using (2), we will get:-
$\tan R = \dfrac{{PQ}}{{\sqrt 3 PQ}} = \dfrac{1}{{\sqrt 3 }}$.
We know that $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$.
Therefore the measure of angle R will be ${30^ \circ }$.

So, the correct answer is “Option D”.

Note: The students might make the mistake of not drawing the triangle and imagining this in their head but that may create chaos and mistakes. So, always try to create diagrams in every question you can.
To approach this question, you just need to focus on the fact that you are just given the sides of a triangle and you have to somehow make it work. So, always just try to minimize the variables.