21N.2.SL.TZ0.5

pestleMathematicsAASLPaper 221N· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

The following diagram shows a semicircle with centre $O$ and radius $r$ . Points $P, Q$ and $R$ lie on the circumference of the circle, such that $PQ = 2 r$ and $R \hat{O} Q = θ$ , where $0 < θ < π$ .

Given that the areas of the two shaded regions are equal, show that $θ = 2 \sin θ$ .

[5]

Hence determine the value of $θ$ .

[1]

Markscheme / solution

attempt to find the area of either shaded region in terms of $r$ and $θ$ (M1)

Note: Do not award M1 if they have only copied from the booklet and not applied to the shaded area.

Area of segment $= \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} \sin θ$ A1

Area of triangle $= \frac{1}{2} r^{2} \sin (π - θ)$ A1

correct equation in terms of $θ$ only (A1)

$θ - \sin θ = \sin (π - θ)$

$θ - \sin θ = \sin θ$ A1

$θ = 2 \sin θ$ AG

Note: Award a maximum of M1A1A0A0A0 if a candidate uses degrees (i.e., $\frac{1}{2} r^{2} \sin (180 ° - θ)$ ), even if later work is correct.

Note: If a candidate directly states that the area of the triangle is $\frac{1}{2} r^{2} \sin θ$ , award a maximum of M1A1A0A1A1.

[5 marks]

$θ = 1.89549 \dots$

$θ = 1.90$ A1

Note: Award A0 if there is more than one solution. Award A0 for an answer in degrees.

[1 mark]

Examiners’ report

[N/A]