A yacht race follows the triangular course shown in the diagram. The course from `P` to `Q` is 1.8 km on a true bearing of 058°.
At `Q` the course changes direction. The course from `Q` to `R` is 2.7 km and `/_PQR = 74^@`.
- What is the bearing of `R` from `Q`? (1 mark)
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- What is the distance from `R` to `P`? (2 marks)
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- The area inside this triangular course is set as a ‘no-go’ zone for other boats while the race is on.
What is the area of this ‘no-go’ zone? (1 mark)
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Show Worked Solution
| i. |  |
`/_ PQS = 58^@ \ \ \ (text(alternate to)\ /_TPQ)`
TIP: Draw North-South parallel lines through relevant points to help calculate angles as shown.
`text(Bearing of)\ R\ text(from)\ Q= 180^@ + 58^@ + 74^@312^@`
ii. `text(Using Cosine rule:)`
| `RP^2` | `=RQ^2+PQ^2- 2 xx RQ xx PQ xx cos/_RQP` |
| `= 2.7^2+1.8^2-2 xx 2.7 xx 1.8 xx cos74^@` | |
| `=7.851…` | |
| `RP` | `= sqrt(7.851…)=2.8019…~~ 2.8\ text(km) (text(1 d.p.))` |
iii. `text(Using)\ A = 1/2 ab sinC:`
| `A` | `= 1/2` `xx 2.7` `xx 1.8` `xx sin74^@` |
| `= 2.3358…` | |
| `= 2.3\ text(km)^2` |
`:.\ text(No-go zone is 2.3 km)^2 .`