In a triangle P Q R , QR is 5cm and angle QPR is 60° calculate the radius of circle of the triangle

May 8, 2026Mathematics

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Answer

\frac{5\sqrt{3}}{3} \text{ cm}

5 steps

Apply the extended Sine Rule, which relates a side of a triangle to the sine of its opposite angle and the circumradius ( $R$ ).
$\frac{a}{\sin A} = 2R$
Here, $a = QR = 5$ cm and $A = \angle QPR = 60^\circ$ .
Substitute the given values into the formula.
$\frac{5}{\sin(60^\circ)} = 2R$
Recall the value of $\sin(60^\circ)$ .
$\sin(60^\circ) = \frac{\sqrt{3}}{2}$
Substitute the sine value and solve for $R$ .
$\frac{5}{\frac{\sqrt{3}}{2}} = 2R$ $5 \times \frac{2}{\sqrt{3}} = 2R$ $\frac{10}{\sqrt{3}} = 2R$ $R = \frac{10}{2\sqrt{3}}$ $R = \frac{5}{\sqrt{3}}$
Rationalize the denominator.
$R = \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$ $R = \frac{5\sqrt{3}}{3}$
The radius of the circumcircle is $\frac{5\sqrt{3}}{3}$ cm.
$R = \boxed{\frac{5\sqrt{3}}{3} \text{ cm}}$

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