Standard +0.3 This is a straightforward application of the sine rule to find the ambiguous case. Students need to recognize that two triangles are possible and find both angles using sin⁻¹ and the supplementary angle. While the ambiguous case requires understanding, the calculation itself is routine for C2 level with clear guidance that two solutions exist.
2. In the triangle \(A B C , A B = 16 \mathrm {~cm} , A C = 13 \mathrm {~cm}\), angle \(A B C = 50 ^ { \circ }\) and angle \(B C A = x ^ { \circ }\) Find the two possible values for \(x\), giving your answers to one decimal place.
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\((\sin x)=\frac{16\times\sin 50}{13}(=0.943 \text{ but accept } 0.94)\)
A1
Give correct value or correct expression for \(\sin x\)
\(x=\text{awrt } 70.5(3) \text{ and } 109.5\) or \(70.6\) and \(109.4\)
dM1 A1
Correct work leading to \(x=\ldots\) via inverse \(\sin\), or correct expression then awrt
Notes:
- M1: Allow awrt 0.77 for \(\sin 50°\)
- A1: If expression given, \(\frac{\sin 50°\times 16}{13}\) is fine. If decimal, allow answers which round to 0.94
- dM1: Correct work leading to \(x=\ldots\) via inverse \(\sin\) in degrees on their value for \(\sin x\) (may need to check on calculator)
- A1: Deduce and state both answers \(x=70.5\) and \(109.5\) (do not need degrees). Accept awrt these. Also accept 70.6 and 109.4
- Special case - Wrong labelling of triangle: Only one solution for angle \(x\). If they find missing side as awrt 12.6 proceed to find an angle or its sine or cosine then give M1A0M0A0
- Alternative Method using cosine rule: M1 uses cosine rule to form three term quadratic equation in \(a\); A1 solves and obtains correct value for \(a\) of awrt 14.6 or awrt 5.95; dM1 correct full method to find (at least) one of the two angles. May use cosine rule again, or find angle \(BAC\) then use sine rule. A1 deduces both correct answers as in main scheme
- NB: Obtaining only one correct angle will usually score M1A1M1A0 in any method
| $\frac{\sin x}{16}=\frac{\sin 50°}{13}$ | M1 | Use sine formula correctly in any form |
| $(\sin x)=\frac{16\times\sin 50}{13}(=0.943 \text{ but accept } 0.94)$ | A1 | Give correct value or correct expression for $\sin x$ |
| $x=\text{awrt } 70.5(3) \text{ and } 109.5$ or $70.6$ and $109.4$ | dM1 A1 | Correct work leading to $x=\ldots$ via inverse $\sin$, or correct expression then awrt |
**Notes:**
- M1: Allow awrt 0.77 for $\sin 50°$
- A1: If expression given, $\frac{\sin 50°\times 16}{13}$ is fine. If decimal, allow answers which round to 0.94
- dM1: Correct work leading to $x=\ldots$ via inverse $\sin$ in degrees on their value for $\sin x$ (may need to check on calculator)
- A1: Deduce and state both answers $x=70.5$ and $109.5$ (do not need degrees). Accept awrt these. Also accept 70.6 and 109.4
- **Special case - Wrong labelling of triangle:** Only one solution for angle $x$. If they find missing side as awrt 12.6 proceed to find an angle or its sine or cosine then give M1A0M0A0
- **Alternative Method using cosine rule:** M1 uses cosine rule to form three term quadratic equation in $a$; A1 solves and obtains correct value for $a$ of awrt 14.6 or awrt 5.95; dM1 correct full method to find (at least) one of the two angles. May use cosine rule again, or find angle $BAC$ then use sine rule. A1 deduces both correct answers as in main scheme
- **NB:** Obtaining only one correct angle will usually score M1A1M1A0 in any method
2. In the triangle $A B C , A B = 16 \mathrm {~cm} , A C = 13 \mathrm {~cm}$, angle $A B C = 50 ^ { \circ }$ and angle $B C A = x ^ { \circ }$ Find the two possible values for $x$, giving your answers to one decimal place.\\
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\hfill \mbox{\textit{Edexcel C2 2017 Q2 [4]}}