Higher June 2018 Paper 2 Q18
18 A triangle has sides of length 8 cm, 10 cm and 14 cm.
Work out the size of the largest angle of the triangle.
Give your answer correct to 1 decimal place.
(3)
| Scheme | Marks |
|---|---|
\(14^2 = 10^2 + 8^2 - 2 \times 10 \times 8 \times \cos A\) or \(\cos A = \dfrac{10^2 + 8^2 - 14^2}{2 \times 8 \times 10}\) oe | M1 |
| M1 | |
| 101.5 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Correct substitution in cosine rule for any angle or for 44.4... or 34.047.... (the other 2 angles to 1dp or better)
M1: \(\cos^{-1}\left(\dfrac{10^2 + 8^2 - 14^2}{2 \times 10 \times 8}\right)\) oe ie \(\cos^{-1}\) of the correct angle or a fully correct method to find the largest angle eg
\(180 - \cos^{-1}\left(\dfrac{196 + 100 - 64}{280}\right) - \cos^{-1}\left(\dfrac{196 + 64 - 100}{224}\right)\) oe
A1: 101.5 to 101.6