Posted by **Sammy** on Sunday, December 2, 2012 at 9:30pm.

How do you get the second triangle when you have an ambiguous case of sine law? For example, if you are given the following information for a triangle:

a = 7.2 mm, b = 9.3 mm, <A = 35°

I can solve for one triangle:

Find <B, <C, side c.

<B:

sinB/9.3 = sin35°/72

sinB = 9.3sin35°/7.2

B = sin^-1 (9.3sin35°/7.2)

B = 48°

<C:

180°-35°-48° = 97°

Side c:

c/sin97° = 7.2/sin35°

c = 7.2/sin97°/sin35°

c = 12.5

How do you find and solve for the second one?

- Math (trigonometry) -
**Reiny**, Sunday, December 2, 2012 at 10:20pm
your calculator gave you 47.80.. or 48°

then angle C = 180-35-48 = 97 , you had that correct

c/sin97 = 7.2/sin35

c = 7.2sin97/sin35

= 12.5

So triangle ABC is such that

a= 7.2 --- given

b = 9.3 -- given

c = 12.5

Angle A = 35 -- given

Angle B = 48

Angle C = 97

but remember the sine is positive in quadrants I or II

so angle B could also have been 180-48 or 132 °

so c/sin48 = 7.2/sin35

c = 7.2sin48/sin35 = 9.3

so angle C = 180-132-35 = 13°

then c/sin13 = 7.2/sin35

c = 7.2sin13/sin35 = 2.8

second case

a=7.2

b=9.3

c= 2.8

A=35°

B = 132°

C = 13°

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