| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring direct application of the cosine rule followed by the area formula (1/2)ab sin C. Part (i) is pure substitution into the cosine rule with simple arithmetic, and part (ii) uses the result from (i) with sin²θ + cos²θ = 1. No problem-solving or insight required—purely routine technique with given values. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
**Question 3(i)**
Use $a^2 = b^2 + c^2 - 2bc\cos A$ but data may be in wrong position **M1**
Obtain $8^2 = 7^2 + 6^2 - 2(7)(6)\cos ABC$ or equivalent **A1**
Derive correctly $\cos ABC = 0.25$ AG **A1**
**Question 3(ii)**
State $\frac{1}{2}ab\sin C$ for the area of a triangle **M1**
Obtain correctly $\sin ABC$ (may be via angle $ABC$ $(= 75.5°)$ or an identity) **M1**
Obtain answers rounding to $20.3$ (cm$^2$) **A1**
**Alternative:**
Use cosine rule to find another angle (angle $A = 46.567$, angle $C = 57.91$) **M1**
Find height of triangle $(5.083)$ **M1**
Use $0.5(\text{base})(\text{height}) = 20.3$ **A1**
**Total: 6 marks**3 A triangle $A B C$ has sides $A B , B C$ and $C A$ of lengths $7 \mathrm {~cm} , 6 \mathrm {~cm}$ and 8 cm respectively.\\
(i) Show that $\cos A B C = \frac { 1 } { 4 }$.\\
(ii) Find the area of triangle $A B C$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2017 Q3 [6]}}