A ship H leaves a port P and sails 30 km due south. Then it sails 60km due west. What is the bearing of H from P?
Let the bearing of H from P be represented by x°.
\(\tan \theta = \frac{30}{60} = 0.5\)
\(\theta = \tan^{-1} (0.5) = 26.565°\)
\(x = 180° + (90 - 26.565)\)
\(x = 180 + 63.435 = 243.435°\)
= \(243° 26'\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}