Given the measures of 2 sides and the included angle of a triangle find the measures of the other angles and the third side of the triangle. Explain how would decide which formula to use, Law of Sines or Law of Cosines. Illustrate with a example drawing. Use side measures 12m and 13m,and angle 32 degrees.
If anyone could help it would be great, because this is the only question on my homework i still cannot seem to understand
If anyone could help it would be great, because this is the only question on my homework i still cannot seem to understand
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Alright. Well first of all, let's say that the 32 degree angle is included in the two sides of 12m and 13m. If we right that another way, it would look a little something like this:
a=12m
b=13m
C=32
(Notice that the lowercase letters represent sides, and the uppercase letters represent angles.)
From here, we can use the law of Cosines to find length c.
c^2=12^2+13^2-2*12*13*cos32
plug that into a calculator, and you get
c=6.96m
Since now we have length c, we can use the law of Sines to determine angle A.
6.96/sin32=12/sinA
Solve and get
A=66.0 degrees
To find angle B, you can add angles C and A together and subtract the sum from 180.
66+32+B=180
B=82
Now, remember that this is how you do it if and only if the angle is included between the two sides. If it isn't, there are other ways to solve it. Hope I helped!
a=12m
b=13m
C=32
(Notice that the lowercase letters represent sides, and the uppercase letters represent angles.)
From here, we can use the law of Cosines to find length c.
c^2=12^2+13^2-2*12*13*cos32
plug that into a calculator, and you get
c=6.96m
Since now we have length c, we can use the law of Sines to determine angle A.
6.96/sin32=12/sinA
Solve and get
A=66.0 degrees
To find angle B, you can add angles C and A together and subtract the sum from 180.
66+32+B=180
B=82
Now, remember that this is how you do it if and only if the angle is included between the two sides. If it isn't, there are other ways to solve it. Hope I helped!