| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Ambiguous case (two solutions) |
| Difficulty | Standard +0.3 This is a standard sine rule ambiguous case question requiring recognition of two possible triangle configurations and straightforward application of formulas. Part (i) uses sine rule with awareness of obtuse angle possibility; part (ii) applies area formula then sine/cosine rule. Slightly above average due to the ambiguous case concept and surd manipulation, but follows textbook patterns closely. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\sin x}{20} = \frac{\sin 45}{16}\) | M1* | Use sine formula correctly in any form. SC B1: \(\frac{\sin x}{16} = \frac{\sin 45}{20}\) |
| \(\sin x = \frac{20\sin 45}{16} \left(= \frac{5\sqrt{2}}{8}\right)\) | A1 | Correct expression for \(\sin x\) or \(0.883\)… |
| Correct work leading to a value for \(x\) | Dep*M1 | If previous A1 not awarded, award for evidence of using inverse sin on their value of \(\sin x\) |
| \(62.1\) and \(117.9\) | A1 [4] | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}(BC)(20)\sin(45) = 75\sqrt{2}\) | M1 | Use \(\frac{1}{2}ab\sin C\) correctly and equate to \(75\sqrt{2}\) |
| \((BC =)\ 15\) (cm) | A1 [2] | Accept 15.0 from correct working |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin x}{20} = \frac{\sin 45}{16}$ | M1* | Use sine formula correctly in any form. SC B1: $\frac{\sin x}{16} = \frac{\sin 45}{20}$ |
| $\sin x = \frac{20\sin 45}{16} \left(= \frac{5\sqrt{2}}{8}\right)$ | A1 | Correct expression for $\sin x$ or $0.883$… |
| Correct work leading to a value for $x$ | Dep*M1 | If previous A1 not awarded, award for evidence of using inverse sin on their value of $\sin x$ |
| $62.1$ and $117.9$ | A1 [4] | Cao |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}(BC)(20)\sin(45) = 75\sqrt{2}$ | M1 | Use $\frac{1}{2}ab\sin C$ correctly and equate to $75\sqrt{2}$ |
| $(BC =)\ 15$ (cm) | A1 [2] | Accept 15.0 from correct working |
---1 In triangle $A B C , A B = 20 \mathrm {~cm}$ and angle $B = 45 ^ { \circ }$.\\
(i) Given that $A C = 16 \mathrm {~cm}$, find the two possible values for angle $C$, correct to 1 decimal place.\\
(ii) Given instead that the area of the triangle is $75 \sqrt { 2 } \mathrm {~cm} ^ { 2 }$, find $B C$.
\hfill \mbox{\textit{OCR PURE Q1 [6]}}