How to solve an acute triangle with a missing side by using the law of cosines.... Segment CB is 45, segment BA is 43, segment CA is unknown "b", <B is 88 degrees. I am supposed to find <A, <C, and segment CA "b"

If I am understanding the problem, we have triangle ABC where CB is 45 units, BA is 43 units, and CA = b is unknown. Also angle at vertex B is 88 degrees. The requirement is to find the other components of the triangle.

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We need some better terminology for the problem:

Let the triangle is ABC, the angles A, B and C, and the sides a, b and c where side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. Now we can restate the problem as:

In triangle ABC, side a is 45 units, side c is 43 units, and angle B is 88 degrees. Find the other components.

The law of cosines is a generalization of the pythagorean theorem that applies to all triangles, not only right triangles.

The law of cosines states that for any side of a triangle b, b

^{2}= a^{2}+ c^{2}-2*a*c*cos(angle B). For any triangle there are three equations like this one, which give a formula to compute each side of a triangle based on the other two sides and the included angle.so b

^{2}= (45)^{2}+ (43)^{2}-2*45*43*cos(88 degrees),and b = sqrt((45)

^{2}+ (43)^{2}-2*45*43*cos(88 degrees)).Now we know the tree sides of the triangle, so we can compute the other two angles with either the Law of Cosines or the Law of Sines.

Example using the law of cosines:

a

^{2}= b^{2}+ c^{2}-2*b*c*cos(angle A),so cos(angle A) = (b

^{2}+ c^{2}- a^{2})/2bcExample using the law of sines:

sin(angle B)/b = sin(angle A)/a, and it follows that

sin(88 degrees)/sqrt((45)

^{2}+ (43)^{2}-2*45*43*cos(88 degrees)) = sin(angle A)/45,so sin(angle A) = 45*sin(88 degrees)/sqrt((45)2 + (43)2 -2*45*43*cos(88 degrees))

You can find angle C the same way.