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### Example A boat travels N29°15' E for 54 miles and then turns N60°45' W for 95 miles. What is the distance and bearing from the starting point? Let θ = α + 29°15' $tan θ = \frac{a}{b} = \frac{95}{54}$ θ = 60°23' α = 60°23' - 29°15' α = 31°08' $sin 60°23' = \frac{95}{C}$ C = $\frac{95}{si...

### Example A boat travels N29°15' E for 54 miles and then turns N60°45' W for 95 miles. What is the distance and bearing from the starting point? Let θ = α + 29°15' $tan θ = \frac{a}{b} = \frac{95}{54}$ θ = 60°23' α = 60°23' - 29°15' α = 31°08' $sin 60°23' = \frac{95}{C}$ C = $\frac{95}{sin 60°23'}$ C = 109.3 mi Distance: 109.3 miles Bearing: N31°08' W