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In this triangle: - AB = 9 cm - AC = 10 cm - BC > 5 cm - angle BCA = 60° - angle ABC < 90° - OCR - GCSE Maths - Question 19 - 2018 - Paper 6

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Question 19

In-this-triangle:---AB-=-9-cm---AC-=-10-cm---BC->-5-cm---angle-BCA-=-60°---angle-ABC-<-90°-OCR-GCSE Maths-Question 19-2018-Paper 6.png

In this triangle: - AB = 9 cm - AC = 10 cm - BC > 5 cm - angle BCA = 60° - angle ABC < 90°. Calculate the area of triangle ABC.

Worked Solution & Example Answer:In this triangle: - AB = 9 cm - AC = 10 cm - BC > 5 cm - angle BCA = 60° - angle ABC < 90° - OCR - GCSE Maths - Question 19 - 2018 - Paper 6

Step 1

Calculate length of side BC

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Answer

Using the Cosine Rule: BC2=AB2+AC2−2imesABimesACimesextcos(BCA)BC^2 = AB^2 + AC^2 - 2 imes AB imes AC imes ext{cos}(BCA) Substituting the known values: BC2=92+102−2imes9imes10imesextcos(60°)BC^2 = 9^2 + 10^2 - 2 imes 9 imes 10 imes ext{cos}(60°) =81+100−90= 81 + 100 - 90 =91= 91 Taking the square root: BC=ext√91=9.54extcm(approx.)BC = ext{√}91 \\ = 9.54 ext{ cm (approx.)}

Step 2

Calculate area of triangle ABC

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Answer

The area can be calculated using the formula: extArea=12×AB×AC×extsin(BCA) ext{Area} = \frac{1}{2} \times AB \times AC \times ext{sin}(BCA) Substituting the known values: extArea=12×9×10×extsin(60°) ext{Area} = \frac{1}{2} \times 9 \times 10 \times ext{sin}(60°) Using the value extsin(60°)=32 ext{sin}(60°) = \frac{\sqrt{3}}{2}: =12×9×10×32= \frac{1}{2} \times 9 \times 10 \times \frac{\sqrt{3}}{2} =22.53extcm2= 22.5\sqrt{3} ext{ cm}^2 Calculating the approximate value: ≈22.5×1.732≈38.89extcm2\approx 22.5 \times 1.732 \approx 38.89 ext{ cm}^2

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