| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Trigonometric identities with triangles |
| Difficulty | Standard +0.8 This question requires translating a geometric constraint (area relationship) into a trigonometric equation, then solving it. Part (a) demands careful setup using the given area condition and the constraint AD=AB, requiring multiple steps of algebraic manipulation. Part (b) involves analyzing solutions graphically or algebraically. While the individual techniques are AS-level standard, the problem-solving insight needed to connect geometry to the trigonometric identity makes this moderately challenging—harder than routine exercises but not requiring advanced techniques. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks |
|---|---|
| 6(a) | Uses area of ADB = area of CDB |
| Answer | Marks | Guidance |
|---|---|---|
| twice | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AC = 2 x AD = 2 x AB or equivalent | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| AC = 2 x AB | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| or equivalent must be justified | AO2.1 | R1 |
| (b)(i) | Uses tan A = sin A |
| Answer | Marks | Guidance |
|---|---|---|
| show two intersections | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| award B2 | AO1.1b | A1 |
| (b)(ii) | Selects A = 60°. (Can be earned | |
| with no other working shown) | AO3.2a | B1 |
| Total | 7 | |
| Q | Alternative marking Instructions | AO |
| 6(a) | Obtains area formula for ABD | |
| using sin A | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| tan A | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| of ABC | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| mathematical argument. | AO2.1 | R1 |
| (b)(i) | Uses tan A = sin A |
| Answer | Marks | Guidance |
|---|---|---|
| show two intersections | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| award B2 | AO1.1b | A1 |
| (b)(ii) | Selects A = 60°. (Can be earned | |
| with no other working shown) | AO3.2a | B1 |
| Total | 7 | |
| Q | Marking Instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Uses area of ADB = area of CDB
or area of ADB = ½ area of ABC
Possibly by use of “½ab sinC”
twice | AO3.1a | M1 | Area of ADB = Area of CDB
CD = AD = c
AC = 2c
𝑐 1
cos A = = so A = 60°
2𝑐 2
tan A = √3
√3
sin A =
2
2sin A = √3 = tan A
Deduces that
AC = 2 x AD = 2 x AB or equivalent | AO1.1b | A1
Uses trigonometry involving sin
and tan based on triangle with
AC = 2 x AB | AO1.1a | M1
Obtains correct conclusion (AG)
Sets out a well-constructed
mathematical argument. Use of 60°
or equivalent must be justified | AO2.1 | R1
(b)(i) | Uses tan A = sin A
cosA
and multiplies
Or Uses sketch of two graphs to
show two intersections | AO1.1a | M1 | sin A = 2 sin A
cos A
sin A = 2 sin A cos A
sin A(1 – 2cos A) = 0
sin A = 0 or cos A = ½
A = 0° and A = 60°
Solves the equation to give A = 0°
and 60°
Or interprets intersections of
graphs of the correct shape between
0° and 90° to be the solutions
Special case
0° and 60° stated but not justified
award B1. Stated and verified
award B2 | AO1.1b | A1
(b)(ii) | Selects A = 60°. (Can be earned
with no other working shown) | AO3.2a | B1 | We need A = 60°
Total | 7
Q | Alternative marking Instructions | AO | Marks | Typical Solution
6(a) | Obtains area formula for ABD
using sin A | AO3.1a | M1 | Area of ADB = ½c2 sin A
BC = tan A
c
BC = c tan A
Area of ABC = ½ c2 tan A
½ c2 tan A = 2 x ½c2 sin A
tan A = 2 sin A
Obtains expression for BC using
tan A | AO1.1a | M1
Obtains correct expression for area
of ABC | AO1.1b | A1
Simplifies to correct conclusion
(AG)
Sets out a well-constructed
mathematical argument. | AO2.1 | R1
(b)(i) | Uses tan A = sin A
cosA
and multiplies
Or Uses sketch of two graphs to
show two intersections | AO1.1a | M1 | sin A = 2 sin A
cos A
sin A = 2 sin A cos A
sin A(1 – 2cos A) = 0
sin A = 0 or cos A = ½
A = 0° and A = 60°
Solves the equation to give A = 0°
and 60°
Or interprets intersections of
graphs of the correct shape between
0° and 90° to be the solutions
Special case
0° and 60° stated but not justified
award B1. Stated and verified
award B2 | AO1.1b | A1
(b)(ii) | Selects A = 60°. (Can be earned
with no other working shown) | AO3.2a | B1 | We need A = 60°
Total | 7
Q | Marking Instructions | AO | Marks | Typical solution$ABC$ is a right-angled triangle.
\includegraphics{figure_6}
$D$ is the point on hypotenuse $AC$ such that $AD = AB$.
The area of $\triangle ABD$ is equal to half that of $\triangle ABC$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan A = 2 \sin A$
[4 marks]
\end{enumerate}
\begin{enumerate}[label=(\alph*), resume]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation given in part (a) has two solutions for $0° \leq A \leq 90°$
[2 marks]
\item State the solution which is appropriate in this context.
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2018 Q6 [7]}}