June 2022 Paper 2 Q16
16 Here is an isosceles triangle.
All the angles are acute.
Not drawn accurately
The area of the triangle is \(120\text{ cm}^2\)
Work out the size of angle \(y\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Uses \(\dfrac{1}{2}ab\sin C\) | ||
| \(\dfrac{1}{2} \times 16 \times 16 \times \sin x\) or \(128\sin x\) | M1 | oe eg \(\dfrac{1}{2} \times 16 \times 16 \times \sin(180 - 2y)\) \(x\) can be any letter or expression may be implied |
| \(\sin x = 120 \div \left(\dfrac{1}{2} \times 16 \times 16\right)\) or \(\sin x = \dfrac{15}{16}\) or \(\sin^{-1}\dfrac{15}{16}\) or \(\sin^{-1}[0.93, 0.94]\) or [68.4, 70.12313] | M1dep | oe eg \(\sin x = \dfrac{240}{256}\) or \(\sin x = [0.93, 0.94]\) equation must have \(\sin x =\) \(x\) can be any letter or expression |
| \(\dfrac{180 - \text{their}[68.4, 70.12313]}{2}\) | M1dep | oe |
| [54.93, 55.8] | A1 | SC2 [75.82, 76.4] |
| Alternative method 2 Works out perpendicular height | ||
| \(120 \div \left(\dfrac{1}{2} \times 16\right)\) or \(120 \div 8\) or 15 | M1 | |
| \(\cos x = \dfrac{\sqrt{16^2 - (\text{their } 15)^2}}{16}\) or \(\cos^{-1}\dfrac{\sqrt{31}}{16}\) or \(\tan x = \dfrac{15}{\sqrt{16^2 - (\text{their } 15)^2}}\) or \(\tan^{-1}\dfrac{15}{\sqrt{31}}\) or [68.4, 70.12313] | M1dep | oe eg \(\sin x = \dfrac{15}{16}\) or \(\sin x = [0.93, 0.94]\) or \(\cos x = [0.34, 0.35]\) or \(\tan x = [2.69, 2.7]\) \(x\) can be any letter or expression |
| \(\dfrac{180 - \text{their}[68.4, 70.12313]}{2}\) | M1dep | oe |
| [54.93, 55.8] | A1 | SC2 [75.82, 76.4] |
| Alternative method 3 Works out perpendicular height | ||
| \(120 \div \left(\dfrac{1}{2} \times 16\right)\) or \(120 \div 8\) or 15 | M1 | oe |
| \(16 - \sqrt{16^2 - (\text{their } 15)^2}\) or \(16 - \sqrt{31}\) or [10.4, 10.44] | M1dep | oe eg \(\tan y = \dfrac{15}{16 - \sqrt{16^2 - (\text{their } 15)^2}}\) \(y\) can be any letter or expression |
| \(\tan^{-1}\dfrac{15}{\text{their}[10.4, 10.44]}\) | M1dep | oe eg \(\tan^{-1}[1.43, 1.44231]\) |
| [54.93, 55.8] | A1 | SC2 [75.82, 76.4] |
Additional guidance
| Alt 1 \(y = [68.4, 70.12313]\) | M1M1 |
| Condone \(\sin =\) for \(\sin x =\) etc Condone \(\sin^{-1} = 0.9375\) for \(\sin^{-1} 0.9375\) etc | |
| SC2 is for omitting the 0.5 from the area of triangle formula | |
| After scoring M1M1, the 3rd M1 is for any full method eg Alt 1 68.6 Cosine rule used to work out the third side of the triangle followed by sine rule to work out \(y\) (up to \(\sin^{-1}\ldots\)) If there are no errors seen in the method the 3rd M1 is awarded and possibly the A1 as well | M1M1 |