A ship H leaves a port P and sails 30 km due south. Then it sails 50km due west. What is the bearing of H from P?
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Correct Answer: Option B
Explanation:
Let the bearing of H from P be represented by x°.
\(\tan \theta = \frac{30}{60} = 0.5\)
\(\theta = \tan^{-1} (0.5) = 26.565°\)
\(x = 180° + (90 - 26.565)\)
\(x = 180 + 63.435 = 243.435°\)
= \(243° 26'\)
Let the bearing of H from P be represented by x°.
\(\tan \theta = \frac{30}{60} = 0.5\)
\(\theta = \tan^{-1} (0.5) = 26.565°\)
\(x = 180° + (90 - 26.565)\)
\(x = 180 + 63.435 = 243.435°\)
= \(243° 26'\)