Law of Cosines: Finding an Angle from Three Sides

The Law of Cosines is a generalization of the Pythagorean theorem to any triangle, not just right ones. It relates the three sides to one of the angles:

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

When $C = 90°$, $\cos C = 0$ and we recover $c^{2} = a^{2} + b^{2}$ — the Pythagorean theorem. For other angles, the $\cos C$ term either subtracts (if $C < 90°$, the triangle is "thinner") or adds (if $C > 90°$, the triangle is "fatter") to the right-angled value.

To solve for an angle given three sides, just rearrange:

$\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2ab}$

Step-by-Step Solution

Given: $a = 7$, $b = 9$, $c = 12$.

Find: Angle $C$ (the angle opposite side $c$).

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Step 1 — Square each side.

$a^{2} = 49 \qquad b^{2} = 81 \qquad c^{2} = 144$

Step 2 — Plug into the rearranged law.

$\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2ab}$

$= \dfrac{49 + 81 - 144}{2(7)(9)}$

$= \dfrac{-14}{126}$

$\approx -0.1111$

Step 3 — Take the inverse cosine.

$C = \arccos(-0.1111) \approx 96.38°$

(In radians: $C \approx 1.682$ rad.)

The negative cosine tells us the angle is obtuse ($C > 90°$). The triangle is "fat" — side $c$ is longer than what a right triangle with legs 7 and 9 would produce ($\sqrt{49 + 81} = \sqrt{130} \approx 11.40$, less than 12).

Step 4 — Find angles $A$ and $B$ for completeness.

Using the Law of Cosines for $A$:

$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc} = \dfrac{81 + 144 - 49}{2(9)(12)} = \dfrac{176}{216} \approx 0.8148$

$A = \arccos(0.8148) \approx 35.43°$

Using $A + B + C = 180°$:

$B = 180° - 35.43° - 96.38° \approx 48.19°$

Step 5 — Verify with the Law of Sines.

Check that $a/\sin A \approx b/\sin B \approx c/\sin C$:

$\dfrac{7}{\sin 35.43°} = \dfrac{7}{0.5798} \approx 12.072$

$\dfrac{9}{\sin 48.19°} = \dfrac{9}{0.7454} \approx 12.075$

$\dfrac{12}{\sin 96.38°} = \dfrac{12}{0.9938} \approx 12.075$

All within rounding error. ✓

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Answer:

$\boxed{\;C = \arccos(-1/9) \approx 96.38°\;}$

The triangle has angles approximately $35.43°,\, 48.19°,\, 96.38°$. Since $C > 90°$, the triangle is obtuse at vertex $C$ — confirmed visually by the wide angle in the visualization.

Try It

• Adjust the three sides with the sliders (subject to the triangle inequality $|b - a| < c < a + b$).
• The triangle redraws in real time and the angles update.
• The HUD shows the Law of Cosines computation step by step.
• Try setting $a^{2} + b^{2} = c^{2}$ (e.g. $a = 3$, $b = 4$, $c = 5$) — angle $C$ should jump to exactly 90°.