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Date May 2008 Marks available 1 Reference code 08M.1.sl.TZ1.7
Level SL only Paper 1 Time zone TZ1
Command term Sketch Question number 7 Adapted from N/A

Question

Triangle \({\text{ABC}}\) is such that \({\text{AC}}\) is \(7{\text{ cm}}\), angle \({\text{ABC}}\) is \({65^ \circ }\) and angle \({\text{ACB}}\) is \({30^ \circ }\).

Sketch the triangle writing in the side length and angles.

[1]
a.

Calculate the length of \({\text{AB}}\).

[2]
b.

Find the area of triangle \({\text{ABC}}\).

[3]
c.

Markscheme

     (A1)     (C1)


Note: (A1) for fully labelled sketch.

[1 mark]

a.

Unit penalty (UP) may apply in this question.

\(\frac{{{\text{AB}}}}{{\sin 30}} = \frac{7}{{\sin 65}}\)     (M1)

(UP)     \({\text{AB}} = 3.86{\text{ cm}}\)     (A1)(ft)     (C2)

Note: (M1) for use of sine rule with correct values substituted.

[2 marks]

b.

Unit penalty (UP) may apply in this question.

\({\text{Angle BAC}} = {85^ \circ }\)     (A1)

\({\text{Area}} = \frac{1}{2} \times 7 \times 3.86 \times \sin {85^ \circ }\)     (M1)

(UP)     \( = 13.5{\text{ }}{{\text{cm}}^2}\)     (A1)(ft)     (C3)

[3 marks]

c.

Examiners report

The triangle was drawn correctly by most and a majority correctly found the length of AB - a few did not write down the units (cm) and so lost a Unit penalty mark. There was still a significant number who tried to use right-angled trigonometry to find the length.

Finding the area of the triangle was mixed with many again assuming the existence of a right angle. Some candidates had their calculators in radian mode rather than degree mode.

a.

The triangle was drawn correctly by most and a majority correctly found the length of AB - a few did not write down the units (cm) and so lost a Unit penalty mark. There was still a significant number who tried to use right-angled trigonometry to find the length.

Finding the area of the triangle was mixed with many again assuming the existence of a right angle. Some candidates had their calculators in radian mode rather than degree mode.

b.

The triangle was drawn correctly by most and a majority correctly found the length of AB - a few did not write down the units (cm) and so lost a Unit penalty mark. There was still a significant number who tried to use right-angled trigonometry to find the length.

Finding the area of the triangle was mixed with many again assuming the existence of a right angle. Some candidates had their calculators in radian mode rather than degree mode.

c.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.0 » Simple two-dimensional shapes and their properties, including perimeters and areas of circles, triangles, quadrilaterals and compound shapes.
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