Thread: A trig problem with two obtuse triangles

It was on a test I wrote, but I just wanna clarify if I did it correctly.

In the diagram,



You're given:

In Triangle ACD,
Angle A, Angle C, and Side CD.

In Triangle ABD,
Angle B and Side AB

You're asked to find Angle A in Triangle ABD.


So, you start off by using the Law of Sines to obtain the value of AD in Triangle ACD.
Once you have that, you use Law of Sines again to obtain Angle D in Triangle ABD.
However, since that angle must be obtuse, you subtract your answer from 180 to obtain the real answer for Angle D.

Now, you just subtract that value and Angle B (given) from 180 to obtain Angle A.

Originally Posted by ty2391

It was on a test I wrote, but I just wanna clarify if I did it correctly.

In the diagram,



You're given:

In Triangle ACD,
Angle A, Angle C, and Side CD.

In Triangle ABD,
Angle B and Side AB

You're asked to find Angle A in Triangle ABD.


So, you start off by using the Law of Sines to obtain the value of AD in Triangle ACD.
Once you have that, you use Law of Sines again to obtain Angle D in Triangle ABD.
However, since that angle must be obtuse, you subtract your answer from 180 to obtain the real answer for Angle D.

Now, you just subtract that value and Angle B (given) from 180 to obtain Angle A.

In triangle ADB the angles satisfy:

DAB=180-ABD-ADB

Also is it stated in the question that angle ADB is obtuse? If not you can't assume it from the
diagram.

RonL

Originally Posted by CaptainBlack

In triangle ADB:

A=180-B-D

RonL

Thank you,

The only thing I wasn't 100% sure about though was the part when I used Law of Sines to get Angle D, but it came out to be an acute angle, so I had to subtract that value from 180 to get the actual value of D.

Was that correct?