Find measures of angles, massive error !

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• Dec 12th 2009, 11:39 AM
Monocerotis
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 ?

http://i45.tinypic.com/dzgy0m.jpg
• Dec 12th 2009, 01:19 PM
bigwave
Quote:

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 ?

http://i45.tinypic.com/dzgy0m.jpg

(angle C + angle A) - 180 = angle B
• Dec 12th 2009, 01:56 PM
Monocerotis
Quote:

Originally Posted by bigwave
(angle C + angle A) - 180 = angle B

and....?
• Dec 12th 2009, 02:28 PM
bigwave
Quote:

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?
• Dec 13th 2009, 10:17 AM
Monocerotis
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 ?
• Dec 13th 2009, 03:22 PM
jgv115
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 $180 - (28.76+10.5) = 140.74$

So $y =140.74$

Now use cosine law to find the right hand side then you can work out x.
• Dec 14th 2009, 02:48 AM
aidan
Quote:

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 ?

http://i45.tinypic.com/dzgy0m.jpg

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

$
\angle CEB = \dfrac{66 \sin(10.5deg)}{25} = 28.7574 deg
$

$\angle CEB = \angle EAB$

$\angle CAB = 180 - 28.7574 = 151.2426
$

$
y = \angle ABD = 180 - (10.5 + 151.2426) = 180 - 161.7426 = 18.2574
$

$x = \angle ADC = 161.7426$

That is what is shown on the post by bigwave.