Use the law of sine

Question

Use the law of sine to solve this problems.

1.Given A=30,a=5, c=8

2.C=75°20' ,a=30 ,c=40

Answer 1

Both of these problems are SSA triangles, so for each problem you have to test for the ambiguous case where you can get 2 triangles to solve, not just one.

 

Indeed, the first is quickly determined to have two triangles.

 

Draw the angle A = 30°.

 

Draw point B on one of A's rays. Label AB as c = 8.

 

Now draw a line perpendicular to the other ray of A that contains point B. That line contains a height of the triangle formed whose length is 8/2 = 4 (it's a 30-60-90 triangle).

 

Notice that since a = 5, 4 < a < 8, which means you can locate point C either to the left or right of the height giving you two triangles to solve.

 

Good luck.

 

In the second problem the calculated height of the right triangle is 38.7 which is greater than a = 30. So a side of length 30 starting at point B will never reach the other ray of A. In this case there is no solution.

Answer 2

Daisy,

 

Law of Sine states: Sin(A)/a=Sin(B)/b=Sin(C)/c

 

In 1, you have A, a and c, you should be able to solve C by Sin(A)/a=Sin(C)/c, from C and A+B+C=180, you should also have B, then use the law of Sine again, you can have b.

 

For 2, it is the same logic. 

 

Hope it helps.

 

Best,

Heng