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Date	May 2022	Marks available	2	Reference code	22M.1.SL.TZ2.3
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Justify and Determine	Question number	3	Adapted from	N/A

Question

The owner of a convenience store installs two security cameras, represented by points $C1$ and $C2$ . Both cameras point towards the centre of the store’s cash register, represented by the point $R$ .

The following diagram shows this information on a cross-section of the store.

The cameras are positioned at a height of $3.1 m$ , and the horizontal distance between the cameras is $6.4 m$ . The cash register is sitting on a counter so that its centre, $R$ , is $1.0 m$ above the floor.

The distance from Camera $1$ to the centre of the cash register is $2.8 m$ .

Determine the angle of depression from Camera $1$ to the centre of the cash register. Give your answer in degrees.

[2]

Calculate the distance from Camera $2$ to the centre of the cash register.

[4]

Without further calculation, determine which camera has the largest angle of depression to the centre of the cash register. Justify your response.

[2]

Markscheme

$\sin θ = \frac{2.1}{2.8}$ OR $\tan θ = \frac{2.1}{1.85202 \dots}$ (M1)

$(θ =) 48.6 ° (48.5903 \dots °)$ A1

[2 marks]

METHOD 1

$\sqrt{2 . 8^{2} - 2 . 1^{2}}$ OR $2.8 \cos (48.5903 \dots)$ OR $\frac{2.1}{\tan (48.5903 \dots)}$ (M1)

Note: Award M1 for attempt to use Pythagorean Theorem with $2.1$ seen or for attempt to use cosine or tangent ratio.

$1.85 (m) (1.85202 \dots)$ (A1)

Note: Award the M1A1 if $1.85$ is seen in part (a).

$(6.4 - 1.85202 \dots)$

$4.55 m (4.54797 \dots)$ (A1)

Note: Award A1 for $4.55$ or equivalent seen, either as a separate calculation or in Pythagorean Theorem.

$\sqrt{{(4.54797 \dots)}^{2} + 2 . 1^{2}}$

$5.01 m (5.00939 \dots m)$ A1

METHOD 2

attempt to use cosine rule (M1)

$(c^{2} =) 2 . 8^{2} + 6 . 4^{2} - 2 (2.8) (6.4) \cos (48.5903 \dots)$ (A1)(A1)

Note: Award A1 for $48.5903 . . . °$ substituted into cosine rule formula, A1 for correct substitution.

$(c =) 5.01 m (5.00939 \dots m)$ A1

[4 marks]

camera $1$ is closer to the cash register (than camera $2$ and both cameras are at the same height on the wall) R1

the larger angle of depression is from camera $1$ A1

Note: Do not award R0A1. Award R0A0 if additional calculations are completed and used in their justification, as per the question. Accept “ $1.85 < 4.55$ ” or “ $2.8 < 5.01$ ” as evidence for the R1.

[2 marks]

Examiners report

Many candidates calculated the angle from vertical rather than the angle of depression.

Candidates could successfully use their vertical angle from (a) or other correct trigonometry, such as Pythagorean theorem or cosine rule, to find the distance from camera 2 to the cash register. This question is a good example of how premature rounding can affect a final answer, and some had an inaccurate final answer because they had rounded intermediate values.

Many provided reasonable justification for their response, even though they often followed correct reasoning with an incorrect conclusion about the larger angle of depression.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig

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Hide related questions

Topic 3—Geometry and trigonometry » SL 3.3—Angles of elevation and depression

Topic 3—Geometry and trigonometry

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Markscheme

Examiners report

Syllabus sections

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