| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Ambiguous case (two solutions) |
| Difficulty | Moderate -0.3 This is an ambiguous case sine rule problem requiring recognition that two triangles are possible, then applying the area formula twice. While it requires understanding of the ambiguous case (a key conceptual point), the execution is straightforward once recognized: use sine rule to find two possible angles at A, then apply area = ½ab sin C. This is slightly easier than average as it's a standard application of a well-known scenario in sine/cosine rule topics, though the ambiguous case does add some conceptual challenge beyond pure routine calculation. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| 1 | 0 |
| 1 | 1 |
| 1 | 2 |
The triangle $A B C$ is such that $A C = 16 \mathrm {~cm} , A B = 25 \mathrm {~cm}$ and $A \widehat { B C } = 32 ^ { \circ }$. Find two possible values for the area of the triangle $A B C$.
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a) Use the binomial theorem to expand $( a + \sqrt { b } ) ^ { 4 }$.\\
b) Hence, deduce an expression in terms of $a$ and $b$ for $( a + \sqrt { b } ) ^ { 4 } + ( a - \sqrt { b } ) ^ { 4 }$.
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a) The vectors $\mathbf { u }$ and $\mathbf { v }$ are defined by $\mathbf { u } = 9 \mathbf { i } - 40 \mathbf { j }$ and $\mathbf { v } = 3 \mathbf { i } - 4 \mathbf { j }$. Determine the range of values for $\mu$ such that $\mu | \mathbf { v } | > | \mathbf { u } |$.\\
b) The point $A$ has position vector $11 \mathbf { i } - 4 \mathbf { j }$ and the point $B$ has position vector $21 \mathbf { i } + \mathbf { j }$. Determine the position vector of the point $C$, which lies between $A$ and $B$, such that $A C : C B$ is $2 : 3$.
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Find the values of $m$ for which the equation $4 x ^ { 2 } + 8 x - 8 = m ( 4 x - 3 )$ has real roots. [5]
\hfill \mbox{\textit{WJEC Unit 1 2018 Q9 [5]}}