Higher November 2021 Paper 1 Q13
13 Here is a triangle \(XYZ\).
Diagram NOT accurately drawn
The length \(XZ\) and the angles \(YXZ\) and \(XYZ\) are each given correct to 2 significant figures.
Calculate the upper bound for the length \(YZ\).
Give your answer correct to one decimal place.
Show your working clearly.
(3)
| Scheme | Marks |
|---|---|
| 15.5 or 16.5 or 24.5 or 25.5 or 125 or 135 | B1 |
| \(\dfrac{(YZ)}{\sin(125)} = \dfrac{16.5}{\sin(24.5)}\) oe | M1 |
| Working required Answer: 32.6 | A1 |
| (3) | |
| (3 marks) |
Notes
B1: Accept
\(16.4\dot{9}\) for 16.5
\(25.4\dot{9}\) for 25.5
\(134.\dot{9}\) for 135
M1: for substitution into sine rule
\(\dfrac{(YZ)}{\sin(LB_2)} = \dfrac{UB_1}{\sin(LB_3)}\) oe where
\(16 \lt UB_1 \leqslant 16.5\) and
\(125 \leqslant LB_2 \lt 130\) and
\(24.5 \leqslant LB_3 \lt 25\)
A1: Accept 32.5 to 32.6 from correct working