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G C S E Triangle

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kuiperbelt | 15:47 Wed 12th Apr 2023 | Science
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Let b = length of AC = 136, c = length of AB and a = length of BC a + b + c = 286, therefore a + c = 286 - b = 286 - 136 = 150 c/a = 2/3 so c = (a + c) × 2/5 = 150 × 2/5 = 60 and a = (a + c) × 3/5 = 150 × 3/5 = 90 The Cosine Rule says c² = a² + b² - 2ab.cos(C) so cos(C) = (a² + b² - c²)/(2ab) = (90² + 136² - 60²)/(90×136) = (8100 + 18496 - 3600)/24480 ≈ 0.939379 Therefore C ≈ cos⁻¹(0.939379) ≈ 20.05°
23:55 Wed 12th Apr 2023
Let b = length of AC = 136, c = length of AB and a = length of BC
a + b + c = 286, therefore a + c = 286 - b = 286 - 136 = 150
c/a = 2/3 so c = (a + c) × 2/5 = 150 × 2/5 = 60 and a = (a + c) × 3/5 = 150 × 3/5 = 90
The Cosine Rule says c² = a² + b² - 2ab.cos(C)
so cos(C) = (a² + b² - c²)/(2ab)
= (90² + 136² - 60²)/(2×90×136)
= (8100 + 18496 - 3600)/24480
≈ 0.939379
Therefore C ≈ cos⁻¹(0.939379) ≈ 20.05°
Nostalgia? The last time I had to use the sine rule or cosine rule, you had to look up values in tables and use logs to do hard sums!

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