Math (trigonometry)
posted by Sammy .
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?

your calculator gave you 47.80.. or 48°
then angle C = 1803548 = 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 18048 or 132 °
so c/sin48 = 7.2/sin35
c = 7.2sin48/sin35 = 9.3
so angle C = 18013235 = 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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