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Date	May 2022	Marks available	2	Reference code	22M.1.SL.TZ1.2
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	2	Adapted from	N/A

Question

A vertical pole stands on horizontal ground. The bottom of the pole is taken as the origin, $O$ , of a coordinate system in which the top, $F$ , of the pole has coordinates $(0, 0, 5.8)$ . All units are in metres.

The pole is held in place by ropes attached at $F$ .

One of the ropes is attached to the ground at a point $A$ with coordinates $(3.2, 4.5, 0)$ . The rope forms a straight line from $A$ to $F$ .

Find the length of the rope connecting $A$ to $F$ .

[2]

Find $FÂO$ , the angle the rope makes with the ground.

[2]

Markscheme

$\sqrt{3 . 2^{2} + 4 . 5^{2} + 5 . 8^{2}}$ (M1)

$= 8.01 (8.00812 \dots) m$ A1

[2 marks]

$FÂO = \sin^{- 1} (\frac{5.8}{8.00812 \dots})$ OR $\cos^{- 1} (\frac{5.52177 \dots}{8.00812 \dots})$ OR $\tan^{- 1} (\frac{5.8}{5.52177 \dots})$ (M1)

$46.4 ° (46.4077 \dots °)$ A1

[2 marks]

Examiners report

This question was the first of its type to be tested and the problem was reduced (erroneously) to 2D trigonometry. Whilst a minority of candidates did tackle this part correctly, many simply arrived at an incorrect length with $\sqrt{{(5.8)}^{2} + {(4.5)}^{2}} = 7.34$ metres.

The incorrect interpretation of the diagram as being 2D, meant that there was an assumption that OA was of length 3.2 metres and so $\tan^{- 1} (\frac{5.8}{3.2}) = 61.1$ ° was seen on many scripts. If, using their incorrect answer for part (a), the candidate had used $\sin^{- 1} (\frac{5.8}{their part (a)})$ it was possible to award “follow through” marks.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints

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Hide related questions

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

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

Topic 3—Geometry and trigonometry

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Markscheme

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