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Determination of Unknown Angles Using Law of Cosines

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Using the Law of Cosines with SSS (to find an angle)
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Sarine draws a triangle. She measures the length of the sides and records her measurements as follows. What is the measure of angle C of the triangle?

a=3\\b=4\\c=5

Guidance

The Law of Cosines, a^2+b^2-2ab \cos C , can be rearranged to facilitate the calculation of the measure of angle C when a, b and c are all known lengths.

a^2+b^2-2ab \cos C &=c^2 \\a^2+b^2-c^2 &=2ab \cos C \\\frac{a^2+b^2-c^2}{2ab} &=\cos C

which can be further manipulated to C=\cos^{-1} \left(\frac{a^2+b^2-c^2}{2ab} \right) .

Example A

Find the measure of the largest angle in the triangle with side lengths 12, 18 and 21.

Solution: First, we must determine which angle will be the largest. Recall from Geometry that the longest side is opposite the largest angle. The longest side is 21 so we will let c = 21 since C is the angle we are trying to find. Let a =12 and b = 18 and use the formula to solve for C as shown. It doesn’t matter which sides we assign to a and b . They are interchangeable in the formula.

m \angle C=\cos^{-1} \left(\frac{12^2+18^2-21^2}{2(12)(18)} \right) \approx 86^\circ

Note: Be careful to put parenthesis around the entire numerator and entire denominator on the calculator to ensure the proper order of operations. Your calculator screen should look like this:

\cos^{-1}((12^2+18^2-21^2)/(2(12)(18)))

Example B

Find the value of x , to the nearest degree.

Solution: The angle with measure x^\circ will be angle C so c = 16, a = 22 and b = 8 . Remember, a and b are interchangeable in the formula. Now we can replace the variables with the known measures and solve.

\cos^{-1} \left(\frac{22^2+8^2-16^2}{2(22)(8)} \right) \approx 34^\circ

Example C

Find the m \angle A , if a = 10, b = 15 and c = 21 .

Solution: First, let’s rearrange the formula to reflect the sides given and requested angle:

\cos A=\left(\frac{b^2+c^2-a^2}{2(b)(c)} \right) , now plug in our values m \angle A=\cos^{-1} \left(\frac{15^2+21^2-10^2}{2(15)(21)} \right) \approx 26^\circ

Concept Problem Revisit

We can use the manipulated Law of Cosines to solve for C.

C=\cos^{-1} \frac{3^2+4^2-5^2}{2(3)(4)} \\C=\cos^{-1} \frac{9 + 16 - 25}{24}\\C=\cos^{-1} \frac{0}{24} = \cos^{-1} 0\\C = 90^\circ

Therefore, the triangle is a right triangle.

Guided Practice

1. Find the measure of x in the diagram:

2. Find the measure of the smallest angle in the triangle with side lengths 47, 54 and 72.

3. Find m \angle B , if a = 68, b = 56 and c = 25 .

Answers

1. \cos^{-1} \left(\frac{14^2+8^2-19^2}{2(14)(8)} \right) \approx 117^\circ

2. The smallest angle will be opposite the side with length 47, so this will be our c in the equation.

\cos^{-1} \left(\frac{54^2+72^2-47^2}{2(54)(72)} \right) \approx 41^\circ

3. Rearrange the formula to solve for m \angle B, \cos B=\left(\frac{a^2+c^2-b^2}{2(a)(c)} \right); \cos^{-1} \left(\frac{68^2+25^2-56^2}{2(68)(25)} \right) \approx 52^\circ

Explore More

Use the Law of Cosines to find the value of x , to the nearest degree, in problems 1 through 6.

  1. Find the measure of the smallest angle in the triangle with side lengths 150, 165 and 200 meters.
  2. Find the measure of the largest angle in the triangle with side length 59, 83 and 100 yards.
  3. Find the m \angle C if a = 6, b = 9 and c=13 .
  4. Find the m \angle B if a = 15, b = 8 and c = 9 .
  5. Find the m \angle A if a = 24, b = 20 and c = 14 .
  6. A triangular plot of land is bordered by a road, a fence and a creek. If the stretch along the road is 100 meters, the length of the fence is 115 meters and the side along the creek is 90 meters, at what angle do the fence and road meet?

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