A motorbike and a car leave a service station at the same time. The motorbike travels on a bearing of 080`@` and the car travles for 15.7km on a bearing on 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far correct to 1 decimal place, has the motorbike travelled?

The answer is 7.7km

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Let S be the position of the service station (please refer to the attached image), B be the position of the bike and C be the position of the car.

Given SC=`15.7` Km. Let SB=`x` Km

`ltBCS=180-(80+28+50)=22^o`

`ltSBC=(80+50)=130^o`

Applying the sine rule:

`x/sin22^o=15.7/sin130^o`

`rArr x=15.7*sin22^o/sin130^o`

`rArr x=7.67~~7.7 Km`

Therefore, the motorbike has travelled a distance of **7.7 Km**.

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How did you get 50 degrees?

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