P, R and Q are points on the circle with centre, O. PT and QT are tangents to the circle at P and Q respectively. ∠PTQ = 60° and ∠OQR = 20°. Writing your reasons clearly, find

1. Identify triangle PTQ as isosceles

Since $PT$ and $QT$ are tangents from the same external point $T$,
$PT = QT.$
Hence $\triangle PTQ$ is isosceles and
$\angle TPQ = \angle TQP.$

2. Use the angle sum of triangle PTQ to find the base angle

In $\triangle PTQ$,
$\angle TPQ + \angle TQP + \angle PTQ = 180^\circ.$
Since $\angle PTQ = 60^\circ$ and $\angle TPQ = \angle TQP$,
$2\angle TQP + 60^\circ = 180^\circ$
$2\angle TQP = 120^\circ$
$\angle TQP = 60^\circ.$

1. Apply the tangent-chord theorem using the answer from part (a)

From part (a), $\angle TQP = 60^\circ$.

Since $QT$ is a tangent and $QP$ is a chord,
$\angle PRQ = \angle TQP = 60^\circ$
by the tangent-chord theorem.

1. Find $\angle ORP$ using the given line $RQ$ and part (b)

From part (b), $\angle PRQ=60^\circ$.

Given $\angle OQR=20^\circ$. Since $OQ=OR$ (radii), triangle $OQR$ is isosceles, so
$\angle ORQ=20^\circ.$
Hence
$\angle ORP=\angle PRQ-\angle ORQ=60^\circ-20^\circ=40^\circ.$

2. Use isosceles triangle $OPR$ to find $\angle OPR$

Since $OP=OR$ (radii), triangle $OPR$ is isosceles.

Thus
$\angle OPR=\angle ORP=40^\circ.$

1. Find the angle at the centre subtended by chord $PQ$

Since $OP \perp PT$ and $OQ \perp QT$,
$\angle OPT=\angle OQT=90^\circ.$
In quadrilateral $OPTQ$,
$\angle POQ=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ.$

2. Find the area of the major sector $QRP$

The angle of the major sector is
$360^\circ-120^\circ=240^\circ.$
Area of major sector
$=\frac{240}{360}\times \pi(5)^2=\frac{50\pi}{3}.$

3. Add the area of triangle $OPQ$ to obtain the major segment

Area of $\triangle OPQ$
$=\frac12(5)(5)\sin120^\circ=\frac{25\sqrt3}{4}.$
Hence area of major segment $QRP$
$=\frac{50\pi}{3}+\frac{25\sqrt3}{4}$
$=63.2\text{ cm}^2 \quad \text{(3 s.f.)}.$