A triangular reading table in the community... | Mathematics Specification B 4MB1 | Paper 1 Mock 01 | Written Paper 1

Question 1 Report

A triangular reading table in the community library is the triangle ABC. \(AB = 1.8\) m, \(AC = 2.4\) m and angle BAC is 74°.

Calculate the length BC, correct to 3 significant figures. (2)
Calculate angle ABC, correct to 1 decimal place. (2)
Find the table area, to 3 significant figures. (1)
Edging strip comes in 7 m lengths. Decide, with a reason, if one length is enough. (1)

Answer

Two sides and the angle between them are known, so the cosine rule is the tool for the third side. The included angle \(BAC\) sits between \(AB\) and \(AC\) and faces the side \(BC\).

(a)

\[BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\] \[= 1.8^2 + 2.4^2 - 2 \times 1.8 \times 2.4 \times \cos 74^{\circ} = 3.24 + 5.76 - 8.64 \times 0.275637...\] \[= 9 - 2.3815... = 6.6185...\]

[1]

\[BC = \sqrt{6.6185...} = 2.5726...\]

so \(BC = 2.57\) m correct to 3 significant figures. [1]

The cosine rule generalises Pythagoras: if the angle were \(90\)° then \(\cos 90^{\circ} = 0\) and the last term would vanish. Because \(74\)° is acute the term is subtracted, making \(BC\) shorter than the \(3\) m that Pythagoras would give.

(b) With a full side and its opposite angle now available, the sine rule is quicker than a second cosine rule. Angle \(ABC\) is opposite \(AC = 2.4\) m, and angle \(BAC = 74\)° is opposite \(BC\):

\[\frac{\sin(ABC)}{2.4} = \frac{\sin 74^{\circ}}{2.5726...} \implies \sin(ABC) = \frac{2.4 \times \sin 74^{\circ}}{2.5726...} = 0.89683...\]

[1]

\[\text{angle } ABC = \sin^{-1}(0.89683...) = 63.734...\]

so angle \(ABC = 63.7\)° correct to 1 decimal place. [1]

The obtuse alternative \(116.3\)° is rejected because it would push the angle sum past \(180\)° when combined with the \(74\)° already known.

(c) Two sides and the included angle again, this time for area:

\[\text{Area} = \frac{1}{2} \times 1.8 \times 2.4 \times \sin 74^{\circ} = 2.16 \times 0.961262... = 2.0763...\]

so the area is \(2.08\) m\(^2\) correct to 3 significant figures. [1]

(d) The edging runs round the perimeter:

\[1.8 + 2.4 + 2.5726... = 6.7726... \text{ m}\]

Since \(6.77\) m is less than \(7\) m, one length is enough. [1]

Use the unrounded \(BC\) in the perimeter. The margin is only about \(0.23\) m, so this is a genuinely close comparison rather than an obvious one.

Read lesson note on Geometrical Properties And Reasoning (PearsonEdexcel)

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