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How to solve an acute triangle with a missing side by using the law of cosines

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"

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Kenneth G. | Experienced Tutor of Mathematics and StatisticsExperienced Tutor of Mathematics and Sta...
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.
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, b2 = a2 + c2 -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 b2=  (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:
a2 = b2 + c2 -2*b*c*cos(angle A),
so cos(angle A) =  (b2 + c2 - a2)/2bc  
Example 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.
Patricia S. | Math Tutoring for K-12 & CollegeMath Tutoring for K-12 & College
5.0 5.0 (39 lesson ratings) (39)
Hi, Haley!
Once you've used the Law of Cosines to solve for another piece of information, there are many different ways to find the remaining missing angle and segment measurements.  Given that you asked about the Law of Cosines in the question title, I will show how to apply the Law of Cosines in each step, but I may use another method to get the answer.
Let's look at a quick recap on the Law of Cosines:  The Law of Cosines states that a2=b2+c2-2b*c*cosA, where a, b, and c are sides of a triangle and A is the interior angle of that triangle that is opposite side a.  There are two other variations of this formula (b2=a2+c2-2a*c*cosB and a2=b2+c2-2b*c*cosA) but you only need to memorize one because you can label the sides of the triangle in any order you want.  The two important things to remember are that (1) the side on the left side of the equation MUST be positioned opposite the angle used with the cosine part of the equation on the right side and (2) the side on the left is only used once in the whole equation - every other side is used twice.
In the question that you gave, CB = 45, BA = 43, and CA = b.  <B = 88º.
By the description (and by how they labeled the sides/angles of the triangle), <B is opposite segment CA (side b), <C is opposite segment BA, and <A is opposite segment CB.  I'm going to start the question by finding CA first, since this is the easiest application of the law of cosines.
a. Find CA.
Law of Cosines:  a2=b2+c2-2b*c*cosA
b=√3738.937 = 61.1468 = 61.1
b. Find <A.
WAY 1: Law of Sines    a    =   b  
sinA      sinB
 CB   =  CA 
sinA      sinB
 45   =  61.1  
sinA      sin(88)
45*sin(88) = 61.1*sinA
44.9726 = 61.1*sinA
0.736 = sinA
<A = 47.3959 = 47.4º
WAY 2: Law of Cosines (again).
(CB)= (BA)+ (CA)- 2BA*CA*cosA
45= 43+ 61.1- 2(43)(61.1)(cosA)
2025 = 1849 + 3733.21 - 5254.6(cosA)
2025 = 5582.21 - 5254.6(cosA)
-3557.21 = -5254.6(cosA)
0.67697 = cosA
<A = 47.3926 = 47.4º
c. Find <C.
WAY 1: The sum of the interior angles of a triangle are 180.
47.4+88+<C = 180
135.4+<C = 180
<C = 44.6º
WAYS 2 & 3: Law of Sines or Law of Cosines (looks similar to part b)
using the Law of Cosines, your equation would look like:
(BA)= (CB)+ (CA)- 2(CB)(CA)cosC
432 = 452 + 61.12 - 2(45)(61.1)cosC
I hope this helps!
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
Triangle ABC: c = 43, a = 45, b = ?, B = 88°.
b^2 = a^2 + c^2 -2ac cos(B)
b^2 = 45^2 + 43^2 -2*45*43*cos(88°)
b ≈ 61.14686376063224
sin(A)/a = sin(B)/b
sin(A) = a/b sin(B)
A = sin-1(a/b sin(B))
A ≈ sin-1(45/61.14686376063224 sin(88°))
A ≈ 47.3481946791299°
C = 180° - (A + B) ≈ 44.6518053208701°