Thread: Find measures of angles, massive error !

So this is the triangle I'm dealing with.

I have to find the measures of the angles x & y.

First thing I tried to do was find the measure of the angle @ A.

a/SinA = c/SinC
66/SinA = 25/Sin10.5

and then I end up with 28 degrees for the angle @ A. Maybe I'm looking at it the wrong way but beyond that I have no idea how to proceed with this question.

Any help ?

Originally Posted by Monocerotis

So this is the triangle I'm dealing with.

I have to find the measures of the angles x & y.

First thing I tried to do was find the measure of the angle @ A.

a/SinA = c/SinC
66/SinA = 25/Sin10.5

and then I end up with 28 degrees for the angle @ A. Maybe I'm looking at it the wrong way but beyond that I have no idea how to proceed with this question.

Any help ?

(angle C + angle A) - 180 = angle B

Originally Posted by bigwave

(angle C + angle A) - 180 = angle B

and....?

Originally Posted by Monocerotis

and....?

if you know angle B (or y) the angle x is 180 - y

I redrew the triangle with a cad program this is what it looks with the angles distances are in feet


is this any help?

The issue here is that I'm getting 28 degrees with the calculation 66/SinA = 25/Sin10.5

the measurement for angle A is obviously not 28 degrees, so basically I'm ending up with an error and I can't understand why because so far as I understand my procedure is correct.

I want to solve for angle A, then because I already have angle C I can solve for angle B.

From angle B I could find the angle next to (x), thereby finding x becasue the sum of the two would be 180.

makes sense now ?

Yes I get 28 as well.

The only thing I can think of is the 10.5 angle is wrong or one of the lengths is wrong.

Doesn't matter, just keep on going.

So

So

Now use cosine law to find the right hand side then you can work out x.

Originally Posted by Monocerotis

So this is the triangle I'm dealing with.

I have to find the measures of the angles x & y.

First thing I tried to do was find the measure of the angle @ A.

a/SinA = c/SinC
66/SinA = 25/Sin10.5

and then I end up with 28 degrees for the angle @ A. Maybe I'm looking at it the wrong way but beyond that I have no idea how to proceed with this question.

Any help ?

Your thinking is correct: the angle at A is 28.7574degrees.
(see my attached image)
But it is NOT the angle CAB, it is the angle EAB or angle CEB












That is what is shown on the post by bigwave.