SPM.1.SL.TZ0.14

pestleMathematicsAISLPaper 1SPM· sl-3-2-2d-and-3d-trigsource ↗

Ollie has installed security lights on the side of his house that are activated by a sensor. The sensor is located at point C directly above point D. The area covered by the sensor is shown by the shaded region enclosed by triangle ABC. The distance from A to B is 4.5 m and the distance from B to C is 6 m. Angle AĈB is 15°.

Find CÂB.

[3]

Point B on the ground is 5 m from point E at the entrance to Ollie’s house. He is 1.8 m tall and is standing at point D, below the sensor. He walks towards point B.

Find the distance Ollie is from the entrance to his house when he first activates the sensor.

[5]

Markscheme / solution

$\frac{{{\text{sin}}\,{\text{C}}\mathop {\text{A}}\limits^ \wedge {\text{B}}}}{6} = \frac{{{\text{sin}}\,15^\circ }}{{4.5}}$ (M1)(A1)

CÂB = 20.2º (20.187415…) A1

Note: Award (M1) for substituted sine rule formula and award (A1) for correct substitutions.

[3 marks]

${\text{C}}\mathop {\text{B}}\limits^ \wedge {\text{D}} = 20.2 + 15 = 35.2^\circ$ A1

(let X be the point on BD where Ollie activates the sensor)

${\text{tan}}\,35.18741 \ldots ^\circ = \frac{{1.8}}{{{\text{BX}}}}$ (M1)

Note: Award A1 for their correct angle ${\text{C}}\mathop {\text{B}}\limits^ \wedge {\text{D}}$ . Award M1 for correctly substituted trigonometric formula.

${\text{BX}} = 2.55285 \ldots$ A1

$5 - 2.55285 \ldots$ (M1)

= 2.45 (m) (2.44714…) A1

[5 marks]

Examiners’ report

[N/A]