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March 30, 2015

March 30, 2015

Posted by **Adam** on Thursday, April 5, 2012 at 11:43pm.

- Trig Help Please!!! -
**Reiny**, Friday, April 6, 2012 at 12:44amYour given information is of the format SAS , so it requires the cosine law to find c

c^2 = 95^2 + 137^2 - 2(95)(137)cos 38°

...

c = 85.335

I would now find angle A using the sine law, then the third angle is easy.

- Trig Help Please!!! -
**Adam**, Friday, April 6, 2012 at 1:21ami got that far too but i cant figure out the angle C!!!!

- Trig Help Please!!! -
**Reiny**, Friday, April 6, 2012 at 9:17amJust noticed that I was actually finding b, not c

(c was given)

No harm done here.

let's find angle A by the sine law

sinA/a = sinB/b

sinA/95 = sin 38/85.335

sinA = .685...

A = 43.27

then angle C = 180-38-43.27 = 97.73° or appr 98°

The problem with the sine law is that we run into the "ambiguous case".

since the sine is positive in I and in II, when we take the inverse sine, we often cannot tell which angle to use.

In this case, since the largest angle is always opposite the largest side, angle C must be the largest.

So I try to avoid finding that angle by the sine law, and I found angle A instead.

Since any triangle could have only ONE obtuse angle, by finding one of the two smaller angles we avoid that problem

- Trig Help Please!!! -
**Amber**, Friday, April 6, 2012 at 12:12pmthank you!!!

- Trig Help Please!!! -
**Laura**, Saturday, May 26, 2012 at 2:50pmHELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

- Trig -
**erin**, Saturday, May 26, 2012 at 2:52pmi dot geht thhis werlk? cahn sumwone elp meh?

- Trig Help Please!!! -
**fFuTQuYeCPRbeDRxAW**, Tuesday, July 24, 2012 at 1:41amI didn't know this, and I feel like I was cheated by my geomtery teacher.I feel like I should have known this I am a math professor with a Ph.D. but I'm comforted to hear that a lot of other people didn't know about it, either.

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