Review question

# How are these three angles connected? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5497

## Solution

The circle in the diagram has centre $C$. Three angles $\alpha, \beta, \gamma$ are also indicated.

The angles $\alpha,\beta, \gamma$ are related by the equation:

1. $\cos \alpha = \sin (\beta + \gamma)$,

2. $\sin \beta = \sin \alpha \sin \gamma$,

3. $\sin \beta (1 − \cos \alpha) = \sin \gamma$,

4. $\sin (\alpha + \beta) = \cos \gamma \sin \alpha$.

Let’s give the vertices in the diagram some names. We’ll call the vertex of angles $\alpha$ and $\beta$, $A$, and the vertex of angle $\gamma$, B.

We’ll also draw in the radius between C and the tangent - they’ll be perpendicular at the point of contact. We’ll call this point $D$.

By looking at the right-angled triangle $ACD$, we can see that the length of $AC = x = \dfrac{r}{\sin \alpha}$, so $r = x\sin\alpha$.

Now we can apply the sine rule to the triangle $ABC$ to give $\frac{\sin \beta}{r}=\frac{\sin \gamma}{x},$ so $x\sin\beta = r\sin\gamma \implies x\sin\beta = x\sin\alpha\sin\gamma.$

So on cancelling $x$ we have $\sin \beta = \sin \alpha \sin \gamma,$ and the answer is (b).