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The side of length "8" is opposite angle C, so it is side c. The other two sides are a and b.

Now let us put what we know into The Law of Cosines:

Start with: |   | c² = a² + b² − 2ab cos(C)
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Put in a, b and c: |   | 8² = 9² + 52 − 2 × 9 × 5 × cos(C)
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Calculate: |   | 64 = 81 + 25 − 90 × cos(C)
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Now we use our algebra skills to rearrange and solve:
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Subtract 25 from both sides: |   | 39 = 81 − 90 × cos(C)
Subtract 81 from both sides: |   | −42 = −90 × cos(C)
Swap sides: |   | −90 × cos(C) = −42
Divide both sides by −90: |   | cos(C) = 42/90
Inverse cosine: |   | C = cos⁻¹(42/90)
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Calculator: |   | C = 62.2° (to 1 decimal place)

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We know angle C = 37º, and sides a = 8 and b = 11

The Law of Cosines says: |   | c² = a² + b² − 2ab cos(C)
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Put in the values we know: |   | c² = 8² + 11² − 2 × 8 × 11 × cos(37º)
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Do some calculations: |   | c² = 64 + 121 − 176 × 0.798…
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More calculations: |   | c² = 44.44...
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Take the square root: |   | c = √44.44 = 6.67 to 2 decimal places

Answer: c = 6.67

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Law of cosines to determine grade | Trig identities and examples | Trigonometry | Khan Academy

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Law of cosines | Trig identities and examples | Trigonometry | Khan Academy

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The law of cosine is used to solve a SAS or SSS triangle in conjuncture with law of sines.
Notes
-When the angle of C is 90 then the formula is the same as the Pythagorean Theorem because the cosine of 90 degrees in 0 which cancels out the latter portion of the equation
-The largest side must be solved first