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Date	November Example questions	Marks available	5	Reference code	EXN.1.AHL.TZ0.7
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	7	Adapted from	N/A

Question

Consider quadrilateral $PQRS$ where $[PQ]$ is parallel to $[SR]$ .

In $PQRS$ , $PQ = x$ , $SR = y$ , $R \hat{S} P = α$ and $Q \hat{R} S = β$ .

Find an expression for $PS$ in terms of $x, y, \sin β$ and $\sin (α + β)$ .

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

METHOD 1

from vertex $P$ , draws a line parallel to $[QR]$ that meets $[SR]$ at a point $X$ (M1)

uses the sine rule in $ΔPSX$ M1

$\frac{PS}{\sin β} = \frac{y - x}{\sin (180 ° - α - β)}$ A1

$\sin (180 ° - α - β) = \sin (α + β)$ (A1)

$PS = \frac{(y - x) \sin β}{\sin (α + β)}$ A1

METHOD 2

let the height of quadrilateral $PQRS$ be $h$

$h = PS \sin α$ A1

attempts to find a second expression for $h$ M1

$h = (y - x - PS \cos α) \tan β$

$PS \sin α = (y - x - PS \cos α) \tan β$

writes $\tan β$ as $\frac{\sin β}{\cos β}$ , multiplies through by $\cos β$ and expands the RHS M1

$PS \sin α \cos β = (y - x) \sin β - PS \cos α \sin β$

$PS = \frac{(y - x) \sin β}{\sin α \cos β + \cos α \sin β}$ A1

$PS = \frac{(y - x) \sin β}{\sin (α + β)}$ A1

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area

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16N.2.SL.TZ0.T_5e:
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