# Find Direction and Angle to arrive directly across initial location?

## Homework Statement

A boat can go 12 km /h in a river that has current of 5 km /h.
At what angle relative to the shore does the boat have to leave to arrive directly across from initial location?

SOH CAH TOA

## The Attempt at a Solution

I just want some explanation of this question. Would my 12km/h be my hypotenuse and the resultant be the straight line across the river perpendicular to the shore or would my 12km/h be the straight line perpendicular to the shore??

Because I get 2 different angles if I use tan-1 (5/12) = 22.16 degrees and sin-1 (5/12) which gives me 24.6 degrees.... Which one do I use?

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jedishrfu
Mentor

No, the 12km/h is the across the river speed and the 5km/h is the river speed so the angle relative to going across the river is tan theta = 5/12. The two speeds add together vectorially giving you a slightly faster speed as you travel across and drift downstream.

No, the 12km/h is the across the river speed and the 5km/h is the river speed so the angle relative to going across the river is tan theta = 5/12. The two speeds add together vectorially giving you a slightly faster speed as you travel across and drift downstream.
I do not understand this, If I want to end up directly across from my initial point, wouldnt the straight line across the river be my resultant and hypotenuse be my 12km/h??

haruspex
Homework Helper
Gold Member

If I want to end up directly across from my initial point, wouldnt the straight line across the river be my resultant and hypotenuse be my 12km/h??
Yes. Consider e.g. what would happen if the boat's speed relative to the water were less than the speed of the current. Now there's no way to get straight across.

Yes. Consider e.g. what would happen if the boat's speed relative to the water were less than the speed of the current. Now there's no way to get straight across.
So, I should use sin -1 (5/12)???

haruspex
Homework Helper
Gold Member

Yes.

jedishrfu
Mentor

I do not understand this, If I want to end up directly across from my initial point, wouldnt the straight line across the river be my resultant and hypotenuse be my 12km/h??
Sorry, I misunderstood your question and was looking puely at the math of adding the velocities.