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In the triangle ABC, AB = 11 cm, BC = 7 cm and CA = 8 cm - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 3

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In the triangle ABC, AB = 11 cm, BC = 7 cm and CA = 8 cm. (a) Find the size of angle C, giving your answer in radians to 3 significant figures. (b) Find the area o... show full transcript

Worked Solution & Example Answer:In the triangle ABC, AB = 11 cm, BC = 7 cm and CA = 8 cm - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 3

Step 1

Find the size of angle C, giving your answer in radians to 3 significant figures.

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Answer

To find angle C in triangle ABC, we can use the cosine rule:

extcos(C)=a2+b2−c22ab ext{cos}(C) = \frac{a^2 + b^2 - c^2}{2ab}

Substituting the values, where a = 11 cm, b = 8 cm, and c = 7 cm:

extcos(C)=112+82−722⋅11⋅8=121+64−49176=136176=0.7727 ext{cos}(C) = \frac{11^2 + 8^2 - 7^2}{2 \cdot 11 \cdot 8} = \frac{121 + 64 - 49}{176} = \frac{136}{176} = 0.7727

Taking the inverse cosine:

C=cos−1(0.7727)≈0.6857 radiansC = \text{cos}^{-1}(0.7727) \approx 0.6857\text{ radians}

Rounding to 3 significant figures:

C≈0.686 radiansC \approx 0.686 \text{ radians}

Step 2

Find the area of triangle ABC, giving your answer in cm² to 3 significant figures.

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Answer

To find the area of triangle ABC, we can use the formula:

extArea=12absin⁡(C) ext{Area} = \frac{1}{2}ab \sin(C)

From part (a), we have found that angle C is approximately 0.686 radians. Now we can use a = 11 cm and b = 8 cm:

extArea=12×11×8×sin⁡(0.686) ext{Area} = \frac{1}{2} \times 11 \times 8 \times \sin(0.686)

Calculating:

sin⁡(0.686)≈0.6343,\sin(0.686) \approx 0.6343,

thus the area becomes:

Area≈12×11×8×0.6343≈27.7cm2.\text{Area} \approx \frac{1}{2} \times 11 \times 8 \times 0.6343 \approx 27.7 cm².

Rounding to 3 significant figures:

Area≈27.7 cm².\text{Area} \approx 27.7 \text{ cm²}.

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