Higher June 2018 Paper 1 Q17
17 Here is a triangle.
Not drawn accurately
Circle the correct equation. [1 mark]
- \(\dfrac{\sin x}{42} = \dfrac{\sin 15^\circ}{104}\)
- \(\dfrac{x}{\sin 42^\circ} = \dfrac{15}{\sin 104^\circ}\)
- \(\dfrac{\sin x}{34} = \dfrac{\sin 15^\circ}{104}\)
- \(\dfrac{x}{\sin 42^\circ} = \dfrac{15}{\sin 34^\circ}\)
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{x}{\sin 42^\circ} = \dfrac{15}{\sin 104^\circ}\) | B1 |