Recipe

Find a triangle's angle from its three side lengths

Use when: you know all three side lengths of a triangle and need the angle opposite one of them, with no angle given directly

Goal: the measure of the angle opposite the side you named, without ever drawing the triangle to scale

Taught in Angles and cosine — the trig this course actually uses

1. 1

   Name the angle you want C, and call the side directly opposite it c. The other two sides get the letters a and b, in either order — it doesn't matter which is which as long as neither is c.

   Example

   A triangle has sides 6, 8, and 9, and you want the angle opposite the side of length 9. Call that angle C, so c = 9; the other two sides are a = 6 and b = 8.

   Why: the Law of Cosines always subtracts the squared side opposite the angle you're solving for — mislabeling which side is c swaps in the wrong number at that step.

2. 2

   Start from c^2 = a^2+b^2-2ab\cos C and isolate \cos C before substituting any numbers: \cos C = \frac{a^2+b^2-c^2}{2ab}

   Example

   With a=6, b=8, c=9, this is the formula you're about to fill in: \cos C = \frac{6^2+8^2-9^2}{2(6)(8)}

   Why: solving for \cos C symbolically first, instead of substituting into c^2=a^2+b^2-2ab\cos C and rearranging afterward, avoids untangling a fraction from inside an equation.

3. 3

   Compute the numerator and denominator separately, then divide to get a decimal.

   Example

   \cos C = \frac{36+64-81}{96} = \frac{19}{96} \approx 0.1979
   • the numerator a^2+b^2-c^2 comes out positive — \cos C>0, so C will be acute (less than 90°).
   • the numerator comes out negative — \cos C<0, so C will be obtuse (more than 90°).

   Why: knowing which sign to expect catches an arithmetic slip before you even take the inverse cosine, rather than after.

4. 4

   Apply the inverse cosine to the decimal from the last step to recover the angle itself.

   Example

   C = \cos^{-1}(0.1979) \approx \boxed{78.6°}

   Why: \cos^{-1} is the one operation that turns a cosine value back into an angle measure — the previous steps only ever produced \cos C, not C.

5. 5

   Compare a^2+b^2 to c^2: if they're equal, the triangle is right and C should land exactly at 90°; otherwise the sign from Step 3 should match whether C came out above or below 90°. Also confirm 0°<C<180°, the only range a triangle's angle can occupy.

   Example

   Here a^2+b^2=100 and c^2=81, so a^2+b^2>c^2 — Step 3 predicted an acute angle, and 78.6° is indeed just under 90°, matching the prediction.

   Why: this catches a mislabeled side or a sign error before the angle gets used anywhere downstream, such as in the dot-product formula this recipe feeds in Chapter 1.