# Math (trigonometry)

posted by on .

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) - ,

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°