angles and sides?

Question

given a=2, beta=20, c=5

have to find b, alpha, gamma

please help me step by step so I can follow, thanks

Answer 1

b^2=a^2+c^2-2accos beta
b^2=4+25-20cos 20
=29-20cos20
29-20*0.9397
=29-18.7939
=10.2061
b=3.195. ANS.
Now sin alpha/a=sinbeta/b
sin alpha=sin20(2/3.195)
=0.2141
alpha=12.36deg.
gamma=180-(20+12.36)=180-32.36=147.65degr.

Answer 2

Use the law of cosines to find b:

b² = a² + c² - 2ac cos(beta)

We know all the quantities on the right side, so plug them in, then solve for b. You should get b ~ 3.195

At this point it might be helpful to draw a sketch of the triangle. Note that c is nearly as much as a+b; so angle gamma is very likely obtuse.

You can use the law of sines or the law of cosines to find the other quantities. It wouldn't hurt to try both, partly for practice and partly to confirm your calculations.

For example, the law of sines gives

sin(alpha)/a = sin(beta)/b so sin(alpha) = a*sin(beta)/b

and the law of cosines gives

a² = b² + c² - 2bc cos(alpha), so

cos(alpha) = (b² + c² - a²)/(2bc)

Both calculations will give you the correct value for alpha (about 12.4 degrees).

Knowing alpha and beta, you can find gamma without using the law of cosines or the law of sines because the sum of the angles of any triangle is 180 degrees. Still, it would be good practice for you to use the laws to find gamma. However, note: the law of sines will give the correct value for the sine of gamma, but if you take the inverse sine of this value, you will not get the correct value for gamma. That's because it turns out that gamma > 90 degrees, and inverse sine always returns a value between -90 degrees and 90 degrees. Using inverse cosine on the result of applying the law of cosines will give the correct value for gamma.

Answer 3

Use the cosine formula to find b, and then the sine formula.

Answer 4

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