| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Given area find angle/side |
| Difficulty | Standard +0.3 This is a straightforward application of the area formula (1/2 ab sin C) to find sin θ, then using sin²θ + cos²θ = 1 to find cos θ, followed by the cosine rule to find BC. The multi-step nature and need to select the correct angle using the constraint elevates it slightly above routine, but all techniques are standard AS-level applications with no novel insight required. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(15 = \frac{1}{2} \times 5 \times 10 \times \sin\theta\) | M1 | Uses Area \(= \frac{1}{2}ab\sin C\) to find \(\sin\theta\) or \(\theta\) |
| \(\sin\theta = \frac{3}{5}\) oe | A1 | May be implied by \(\theta =\) awrt \(36.9°\) or awrt \(0.644\) radians |
| Uses \(\cos^2\theta = 1 - \sin^2\theta\) | M1 | Uses their \(\sin\theta\) to find two values of \(\cos\theta\). Also allow triangle method or graphical calculator. Values must be symmetrical \(\pm k\) |
| \(\cos\theta = \pm\frac{4}{5}\) | A1 | Or \(\pm 0.8\). Condone values appearing from \(\pm 0.79...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(BC^2 = 10^2 + 5^2 - 2\times10\times5\times\text{"}-\frac{4}{5}\text{"}\) | M1 | Uses negative value of \(\cos\theta\) (obtuse angle) to find \(BC\) using cosine rule. Alternatively works out \(BC\) using both values and chooses larger. If cosine rule stated it must have minus sign. Using \(+\)ve acute angle is M0 A0 |
| \(BC = \sqrt{205}\) | A1 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $15 = \frac{1}{2} \times 5 \times 10 \times \sin\theta$ | M1 | Uses Area $= \frac{1}{2}ab\sin C$ to find $\sin\theta$ or $\theta$ |
| $\sin\theta = \frac{3}{5}$ oe | A1 | May be implied by $\theta =$ awrt $36.9°$ or awrt $0.644$ radians |
| Uses $\cos^2\theta = 1 - \sin^2\theta$ | M1 | Uses their $\sin\theta$ to find two values of $\cos\theta$. Also allow triangle method or graphical calculator. Values must be symmetrical $\pm k$ |
| $\cos\theta = \pm\frac{4}{5}$ | A1 | Or $\pm 0.8$. Condone values appearing from $\pm 0.79...$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $BC^2 = 10^2 + 5^2 - 2\times10\times5\times\text{"}-\frac{4}{5}\text{"}$ | M1 | Uses negative value of $\cos\theta$ (obtuse angle) to find $BC$ using cosine rule. Alternatively works out $BC$ using both values and chooses larger. If cosine rule stated it must have minus sign. Using $+$ve acute angle is M0 A0 |
| $BC = \sqrt{205}$ | A1 | |
---\begin{enumerate}
\item In a triangle $A B C$, side $A B$ has length 10 cm , side $A C$ has length 5 cm , and angle $B A C = \theta$ where $\theta$ is measured in degrees. The area of triangle $A B C$ is $15 \mathrm {~cm} ^ { 2 }$\\
(a) Find the two possible values of $\cos \theta$
\end{enumerate}
Given that $B C$ is the longest side of the triangle,\\
(b) find the exact length of $B C$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q7 [6]}}