SOLUTION: the three given points are the vertices of a triangle. solve the triangle, rounding lengths tot he nearest tenth and angle measures to the nearest degree A(0,0) B(-3,5) C(3,-2)

Question 929573: the three given points are the vertices of a triangle. solve the triangle, rounding lengths tot he nearest tenth and angle measures to the nearest degree A(0,0) B(-3,5) C(3,-2)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
the three given points are the vertices of a triangle. solve the triangle, rounding lengths tot he nearest tenth and angle measures to the nearest degree A(0,0) B(-3,5) C(3,-2)
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Plot given points to form a triangle with angle A at (0,0), B at (-3,5) and C at (3,-2).
Using distance formula:
AB=√(0+3)^2+(0-5)^2)=√(9+25)=√34
AC=√(0-3)^2+(0+2)^2)=√(9+4)=√13
BC=√(-3-3)^2+(5+2)^2)=√(36+49)=√85
..
Law of cosine: c^2=a^2+b^2-2abcosA
(BC)^2=(AB)^2+(AC)^2-2*AB*ACcosA
√85^2=34+13-2*√34*√13cosA
85=47-2√442cosA
38=-42cosA
cosx=-38/42
A=154.8˚
..
using law of sin:
BC/sin(154.8˚)=AC/sinB
sinB=AC*sin(154.8˚)/BC=0.1665
B=9.6˚
..
BC/sin(154.8˚)=AB/sinC
sinC=AB*sin(154.8˚)/BC=0.2693
C=15.6˚

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