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1. The diagram below shows a triangle ABC, where AB = 8 cm, AC = 6 cm, and angle ACB = 60 degrees - HSC - SSCE Mathematics Advanced - Question 1 - 2020 - Paper 1

Question 1

1. The diagram below shows a triangle ABC, where AB = 8 cm, AC = 6 cm, and angle ACB = 60 degrees. (Insert diagram here) (a) Calculate the length of side BC.... show full transcript

Worked Solution & Example Answer:1. The diagram below shows a triangle ABC, where AB = 8 cm, AC = 6 cm, and angle ACB = 60 degrees - HSC - SSCE Mathematics Advanced - Question 1 - 2020 - Paper 1

Step 1

Calculate the length of side BC.

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Answer

To find the length of side BC, we will apply the Law of Cosines, which states:

$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$

In our case:

Let AB = c = 8 cm,
AC = b = 6 cm,
angle ACB = C = 60 degrees,
BC = a (unknown).

Inserting these values into the formula gives: $BC^2 = 8^2 + 6^2 - 2 \cdot 8 \cdot 6 \cdot \cos(60^{\circ})$ $BC^2 = 64 + 36 - 2 \cdot 8 \cdot 6 \cdot \frac{1}{2}$ $BC^2 = 100 - 48$ $BC^2 = 52$ $$BC = \sqrt{52} \approx 7.21 \text{ cm}.$

Step 2

Calculate the area of triangle ABC.

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Answer

The area of triangle ABC can be calculated using the formula:

$\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin(C)$

Substituting the known values:

a = 8 cm,
b = 6 cm,
angle C = 60 degrees.

Thus, the area becomes: $\text{Area} = \frac{1}{2} \cdot 8 \cdot 6 \cdot \sin(60^{\circ})$ $\text{Area} = \frac{1}{2} \cdot 8 \cdot 6 \cdot \frac{\sqrt{3}}{2}$ $$\text{Area} = 24\sqrt{3} \approx 41.57 \text{ cm}^2.$

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