Standard +0.3 This is a straightforward application of the sine rule to find a side in triangle ABJ, followed by using trigonometry to find the perpendicular distance (width). The setup is clear with given angles and one side, requiring 2-3 standard steps with no conceptual challenges beyond recognizing which trigonometric approach to use.
6 A and B are points on the same side of a straight river. A and B are 180 metres apart. The angles made with a jetty J on the opposite side of the river \(78 ^ { \circ }\) and \(56 ^ { \circ }\) respectively as shown.
\includegraphics[max width=\textwidth, alt={}, center]{c55a5f04-3573-4f36-a12c-3755bdd4a45b-3_332_681_1451_565}
Not to scale
Calculate the width of the river correct to the nearest metre.
Use of sine rule with 46° \(\frac{180}{\sin 46°} = \frac{JB}{\sin 78°}\) JB = 244.76
M1 A1 A1
Use of right angled triangle \(\sin 56° = \frac{\text{width}}{244.76}\) width = 203 m
Use of sine rule with 46° $\frac{180}{\sin 46°} = \frac{JB}{\sin 78°}$ JB = 244.76 | M1 A1 A1 | Use of right angled triangle $\sin 56° = \frac{\text{width}}{244.76}$ width = 203 m | M1 A1 | 5 marks | Alternatively work with JA
6 A and B are points on the same side of a straight river. A and B are 180 metres apart. The angles made with a jetty J on the opposite side of the river $78 ^ { \circ }$ and $56 ^ { \circ }$ respectively as shown.\\
\includegraphics[max width=\textwidth, alt={}, center]{c55a5f04-3573-4f36-a12c-3755bdd4a45b-3_332_681_1451_565}
Not to scale
Calculate the width of the river correct to the nearest metre.
\hfill \mbox{\textit{OCR MEI C2 Q6 [5]}}