Geometry & Right Triangle Trig

[asissa]

Okay, here is the word problem: A surveyor needs to find the distance BC across a lake as part of a project to build a bridge. The distance from point A to point B is 325 ft. The measurement of angle A is 42 degrees and the measurement of angle B is 110 degrees. what is the distance BC across the lake to the nearest foot? The answer in the back of the book is 463ft. I get every number but that! Please let me know where I've gone wrong.

I started with the cos 42 = 325/hyp.
or, .7431 = 325/hyp
or, 437.35

Then, sin 42 = opposite/437.35
or, .6691 = opposite/437.35
or, .6691 x 437.35
or 292.63

Where is my error?
Thank you!

 

[asissa]

I just realized I don't have a right triangle. Do I put a line through the 110 degree angle and then solve for one triangle to get the other?

 

[Deleted member 4993]

Okay, here is the word problem: A surveyor needs to find the distance BC across a lake as part of a project to build a bridge. The distance from point A to point B is 325 ft. The measurement of angle A is 42 degrees and the measurement of angle B is 110 degrees. what is the distance BC across the lake to the nearest foot? The answer in the back of the book is 463ft. I get every number but that! Please let me know where I've gone wrong.

I started with the cos 42 = 325/hyp. <<< Where is the hyp.? Where is right angled triangle?
or, .7431 = 325/hyp
or, 437.35 <<<< What's that

Then, sin 42 = opposite/437.35
or, .6691 = opposite/437.35
or, .6691 x 437.35
or 292.63

Where is my error?
Thank you!

Have you worked with laws of Sines in a triangle - where it states:

\(\displaystyle \frac{a}{sinA}\ = \ \frac{b}{sinB}\ = \ \frac{c}{sinC}\\)

 

[asissa]

Thank you for your help! I just figured it out - I did need to divide the triangle into two right triangles (through the 110 degree angle). Angle C is 28 degrees. By doing the sin 42 and then using that answer to get the sin of 28, I get 463.