Trig Problem (1 Viewer)

[Nathan Truong]

Hello,

Here is the question:

Maryagok was standing on the edge of a cliff, where she spotted two pirate ships at angles of depression of 63° and 41°. If the distance between herself and the ship closer to the cliff was 520m.

Find the distance between herself and the ship further away from the cliff (3dp).

Thanks in advance.

 

[RealiseNothing]

Let's call the height of the cliff 'y' and try to find it.

We can see from the triangle made from the girl, base of the cliff, and the closest ship that cos(27) = y/520

Rearranging gives y = 520cos(27)

Let's leave it like that since that's it's exact form.

Now to find the distance between herself and the second ship, we can now use cos(49) = y/x where I've let x = distance to the second ship.

Rearranging gives x = y/cos(49)

Sub in our y value:

x = 520cos(27) / cos(49)

x = 706.222 to 3 decimal places.

 

[Aysce]

It should be tan(49) = y/520 ^

Woops nvm

 

[RealiseNothing]

It should be tan(49) = y/520 ^

It's the distance from herself to the boat is 520, so the hypotenuse is 520, not the base of the triangle. I did what you did at first before realising this.

 

[Aysce]

Yeah, misread I suppose -.-