Find Direction and Angle to arrive directly across initial location?

[Stanc]

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?

Homework Equations

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?

 

[jedishrfu]

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.

 

[Stanc]

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, wouldn't the straight line across the river be my resultant and hypotenuse be my 12km/h??

 

[haruspex]

If I want to end up directly across from my initial point, wouldn't 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.

 

[Stanc]

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]

Yes.

 

[jedishrfu]

I do not understand this, If I want to end up directly across from my initial point, wouldn't 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.