Angles in Semicircles

Proving the Right Angle

To prove a mathematical rule is universally true, we cannot just measure one triangle; we must use algebra and fundamental geometric properties.

• Imagine a triangle $ABC$ inscribed inside a circle, where $AB$ is the diameter passing through the centre $O$, and $C$ is a point on the edge.
• If we draw a radius from the centre $O$ to the point $C$, we split the large triangle into two smaller triangles: $\triangle AOC$ and $\triangle BOC$.
• The lines $OA$, $OB$, and $OC$ are all exactly the same length because they are all radii of the same circle.
• This means our two smaller triangles are each an isosceles triangle.

Because isosceles triangles have equal base angles, we can assign algebraic values to them to prove the theorem:

1. Let $\angle OAC = \angle OCA = x$.
2. Let $\angle OBC = \angle OCB = y$.

We know the sum of all angles in the large triangle ($\triangle ABC$) must be $180^\circ$. We can write this as an equation and simplify it:

$$x + y + (x + y) = 180^\circ$$ $$2x + 2y = 180^\circ$$ $$2(x + y) = 180^\circ$$ $$x + y = 90^\circ$$

Since the angle at $C$ is equal to $x + y$, we have successfully proven that the angle is exactly $90^\circ$.

Calculating Missing Angles and Sides

Spotting this hidden $90^\circ$ angle is often the crucial first step to unlocking complex multi-step circle geometry problems.

• Once you identify the $90^\circ$ angle, you know that the remaining two angles in the triangle must also add up to $90^\circ$ (since $180^\circ - 90^\circ = 90^\circ$).

Triangle $PQR$ is drawn inside a circle with diameter $PR$. If $\angle QPR = 41^\circ$, calculate $\angle PRQ$. Give reasons for your answer.

Step 1: Identify the right angle using the circle theorem.

• $\angle PQR = 90^\circ$ (Reason: Angle in a semicircle is $90^\circ$)

Step 2: Use the sum of angles in a triangle to find the missing angle.

• $\angle PRQ = 180^\circ - (90^\circ + 41^\circ)$

Step 3: Calculate the final value.

• $\angle PRQ = 49^\circ$

A right-angled triangle is inscribed in a semicircle. The diameter of the circle is $17\text{ cm}$ and one of the shorter chords is $8\text{ cm}$. Calculate the length of the other chord, $x$.

Step 1: State the formula for Pythagoras' theorem, recognizing the diameter is the hypotenuse ($c$).

• $a^2 + b^2 = c^2$

Step 2: Substitute the known values.

• $x^2 + 8^2 = 17^2$
• $x^2 + 64 = 289$

Step 3: Rearrange and solve for $x$.

• $x^2 = 225$
• $x = 15\text{ cm}$