Question

In triangle PQR, right angled at Q, PQ = 3cm and PR = 6cm. Determine angle P and R.

Answer 1

Hint: First we draw the diagram and plot the given information in it and then we will use the fact that the sum of all the angles in a triangle is ${{180}^{\circ }}$ , and we will also use the sin formula $\dfrac{\sin P}{p}=\dfrac{\sin Q}{q}=\dfrac{\sin R}{r}$ to find the angles P and R.

Complete step-by-step answer:
First of all we will draw a trigonometric standard angles table which is as follows:

Degrees	${{0}^{\circ }}$	${{30}^{\circ }}$	${{45}^{\circ }}$	${{60}^{\circ }}$	${{90}^{\circ }}$
Radians	0	$\dfrac{\pi }{6}$	$\dfrac{\pi }{4}$	$\dfrac{\pi }{3}$	$\dfrac{\pi }{2}$
Sine	0	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$	1
Cosine	1	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$	0
Tangent	0	$\dfrac{1}{\sqrt{3}}$	1	$\sqrt{3}$	Not defined

Let’s start our solution,

In the above figure,
q = PR = 6cm, p = RQ and r = PQ = 3cm
Now we will use the Pythagoras theorem to find the side RQ or p,
The formula is: ${{p}^{2}}={{q}^{2}}-{{r}^{2}}$
Now substituting the value of q and r we get,
$\begin{align}
  & {{p}^{2}}={{6}^{2}}-{{3}^{2}} \\
 & p=\sqrt{36-9}=3\sqrt{3} \\
\end{align}$
Now the sin formula states that $\dfrac{\sin P}{p}=\dfrac{\sin Q}{q}=\dfrac{\sin R}{r}$ must be true,
Therefore, using the above formula we get,
$\dfrac{\sin P}{p}=\dfrac{\sin Q}{q}$
Now substituting the values of $p=3\sqrt{3}$ , Q = ${{90}^{\circ }}$ , q = 6 we get,
$\begin{align}
  & \dfrac{\sin P}{3\sqrt{3}}=\dfrac{\sin {{90}^{\circ }}}{6} \\
 & \sin P=\dfrac{\sqrt{3}}{2} \\
\end{align}$
Hence, P = ${{60}^{\circ }}$ .
Now we know that the sum of all the angles in a triangle is 180, using this we get,
$\angle P+\angle Q+\angle R={{180}^{\circ }}$
Now substituting P = ${{60}^{\circ }}$ and Q = ${{90}^{\circ }}$ we get,
$\angle R={{180}^{\circ }}-{{90}^{\circ }}-{{60}^{\circ }}={{30}^{\circ }}$
Hence we have found the value of angle P and R.

Note: The cosine and the sin formula of the triangles are very important and used to solve this type question very easily. One can also use the cosine formula $\cos P=\dfrac{{{q}^{2}}+{{r}^{2}}-{{p}^{2}}}{2qr}$ , and the substitute all the given values to find the angle P. Hence, all these formulas must be kept in mind.