# I dont understand this triangle math question

• April 4th 2009, 09:20 PM
yeunju
I dont understand this triangle math question
If its not too much trouble

The Directions said: two triangles can be formed using the given measurements. Solve both triangles.

A= 60 degrees, side a = 12, side b = 13

I drew the triangle, found all sides and angles, but i dont get how "two triangles can be formed" (i cant visualize it, because i dont know how to draw it with 2 triangles?)
• April 4th 2009, 10:16 PM
Triangles: the Ambiguous Case
Hello yeunju
Quote:

Originally Posted by yeunju
If its not too much trouble

The Directions said: two triangles can be formed using the given measurements. Solve both triangles.

A= 60 degrees, side a = 12, side b = 13

I drew the triangle, found all sides and angles, but i dont get how "two triangles can be formed" (i cant visualize it, because i dont know how to draw it with 2 triangles?)

A question like this, where you are given two sides of a triangle and the non-included angle, and two triangles are possible, is called the Ambiguous Case. The attached diagram shows how you can draw the two triangles with the measurements you were given.

You've probably found the one where $\angle B$ is acute (that's $B_2$ in my diagram). But there's also a possibility that $\angle B$ is obtuse (like $B_1$). Since $\triangle B_1CB_2$ is isosceles, you can probably see why $\angle AB_1C = 180^o - \angle AB_2C$.

Can you complete the solution of the second triangle now?

• April 5th 2009, 11:02 AM
yeunju
Triangles
So the direction is :
Two triangles can be formed using the given measurements. Solve both triangles.

angle A= 60 degrees, side a = 12, side b = 13

I used the law of sines to find the rest of the angles and sides

Angle B= 70 degrees, angle c= 50 degrees, side c = 10.6 (i drew it with c as the base)

My issue was knowing how to form two triangles with the measurements. I drew the triangle, found all the sides and angles..but i dont understand forming two triangles?

My work to get the rest of the measurements:

sin(60)/12 = sin(x)/side 13 (cross multiplied)
x= angle b = 69.75 about 70 degrees

180-(angle B+angle A)= 180-(70+60) = 50 degrees (angle C)

Sin 60/side 12 = sin50/ x (cross multiplied)
x = side c = 10.6
• April 5th 2009, 11:11 AM
red_dog
$\sin x=\sin(180-x), \ x\in(0,90)$

So, $B=x$ or $B=180-x$