AQA AS Paper 2 2019 June — Question 5 4 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeReal-world application problems
DifficultyModerate -0.3 This is a straightforward application of sine rule to find the height of the triangle, followed by basic area and volume formulas. The question requires multiple steps (find angle BAC, use sine rule for AB, calculate area, multiply by length) but each step is routine and clearly signposted. It's slightly easier than average because it's a standard textbook-style problem with no conceptual challenges or problem-solving insight required.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)
5 A triangular prism has a cross section \(A B C\) as shown in the diagram below. Angle \(A B C = 25 ^ { \circ }\) Angle \(A C B = 30 ^ { \circ }\) \(B C = 40\) millimetres. The length of the prism is 300 millimetres.
Calculate the volume of the prism, giving your answer to three significant figures.
Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{AB}{\sin 30} = \frac{40}{\sin 125}\), uses sine rule with 125° (or 55°)M1 (AO 1.1a)
\(AB = \frac{40\sin 30}{\sin 125} = 24.415\), finds one side; \(AB = 24.4\) or \(AC = 20.6\)A1 (AO 1.1b)
Area \(= \frac{1}{2} \times 40 \times \frac{40\sin 30}{\sin 125} \times \sin 25\), uses \(\frac{1}{2}ab\sin C\)M1 (AO 1.1a) NB must be a valid set
\(= 206.4\); Volume \(= 61900 \text{ mm}^3\)A1 (AO 1.1b) Condone missing units
Alternative Solution Q5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Uses \(\tan 30°\) and \(\tan 25°\) separately to obtain expressions for vertical height; \(h = x\tan 30\), \(h = (40-x)\tan 25\)M1 (AO 1.1a)
\(h + \frac{h}{\tan 30}\tan 25 = 40\tan 25\); obtains correct expression for \(h\); \(h = \frac{40\tan 25}{(1+\frac{\tan 25}{\tan 30})} = 10.3184...\)A1 (AO 1.1b) PI by correct area
Area \(= \frac{1}{2} \times 40 \times 10.3184... = 206.4\); uses \(\frac{1}{2} \times \text{base} \times \text{height}\)M1 (AO 1.1a) Must see a calculated height
Volume \(= 61900 \text{ mm}^3\)A1 (AO 1.1b) Condone missing units 
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{AB}{\sin 30} = \frac{40}{\sin 125}$, uses sine rule with 125° (or 55°) | M1 (AO 1.1a) | |
| $AB = \frac{40\sin 30}{\sin 125} = 24.415$, finds one side; $AB = 24.4$ or $AC = 20.6$ | A1 (AO 1.1b) | |
| Area $= \frac{1}{2} \times 40 \times \frac{40\sin 30}{\sin 125} \times \sin 25$, uses $\frac{1}{2}ab\sin C$ | M1 (AO 1.1a) | NB must be a valid set |
| $= 206.4$; Volume $= 61900 \text{ mm}^3$ | A1 (AO 1.1b) | Condone missing units |

**Alternative Solution Q5:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $\tan 30°$ and $\tan 25°$ separately to obtain expressions for vertical height; $h = x\tan 30$, $h = (40-x)\tan 25$ | M1 (AO 1.1a) | |
| $h + \frac{h}{\tan 30}\tan 25 = 40\tan 25$; obtains correct expression for $h$; $h = \frac{40\tan 25}{(1+\frac{\tan 25}{\tan 30})} = 10.3184...$ | A1 (AO 1.1b) | PI by correct area |
| Area $= \frac{1}{2} \times 40 \times 10.3184... = 206.4$; uses $\frac{1}{2} \times \text{base} \times \text{height}$ | M1 (AO 1.1a) | Must see a calculated height |
| Volume $= 61900 \text{ mm}^3$ | A1 (AO 1.1b) | Condone missing units |

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5 A triangular prism has a cross section $A B C$ as shown in the diagram below.

Angle $A B C = 25 ^ { \circ }$\\
Angle $A C B = 30 ^ { \circ }$\\
$B C = 40$ millimetres.

The length of the prism is 300 millimetres.\\
Calculate the volume of the prism, giving your answer to three significant figures.\\

\hfill \mbox{\textit{AQA AS Paper 2 2019 Q5 [4]}}
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