# Angles in a semicircle

When a triangle is formed inside a semicircle, two lines from either side of the diameter meet at a point on the circumference at a right angle.

The angle in a semicircle is a right angle of $$90^\circ$$.

Question

In the diagram PR is a diameter and $$\angle PRQ = 25^\circ$$.

What is the size of $$\angle QPR$$?

$$\angle PQR = 90^\circ$$ since it is the angle in a semicircle.

The three angles in the triangle add up to $$180^\circ$$, therefore:

$\angle QPR = 180^\circ - 90^\circ - 25^\circ$

$\angle QPR = 65^\circ$

Question

In the diagram KL is a diameter of the circle and is 8 cm long.

LM = 3 cm.

Calculate the size of KM.

KL is a diameter so we have an angle in a semicircle therefore $$\angle KML = 90^\circ$$.

We have a right-angled triangle and so can use Pythagoras.

KM is not the hypotenuse so:

$K{M^2} = {8^2} - {3^2} = 55$

$KM = \sqrt {55} = 7.461...$

$KM = 7.4cm\,(to\,1\,d.p.)$