How to Calculate the Third Side of a Triangle
Finding the length of the missing side of a triangle is a fundamental skill in geometry, trigonometry, and many real‑world applications such as engineering, architecture, and navigation. That said, whether you are dealing with a right triangle, an oblique triangle, or a situation where only partial information is known, there are reliable formulas that let you determine the unknown side with confidence. This guide walks you through the most common scenarios, explains the underlying mathematics, and provides step‑by‑step examples so you can apply the techniques yourself And that's really what it comes down to..
Understanding Triangle Basics
Before diving into calculations, it helps to refresh the basic elements of a triangle:
- Sides: Usually denoted a, b, and c.
- Angles: Opposite the sides, denoted A, B, and C (angle A is opposite side a, etc.).
- Known vs. Unknown: In most problems you will know two sides and either the included angle or an opposite angle, or you will know two angles and one side.
The triangle inequality theorem states that any side must be shorter than the sum of the other two sides and longer than their difference. This rule is a quick sanity check after you compute a missing length.
Methods to Calculate the Third Side
Depending on what information you have, you will use one of the following approaches:
| Known Information | Appropriate Formula | When to Use |
|---|---|---|
| Two sides and the included angle (SAS) | Law of Cosines: c² = a² + b² – 2ab·cos(C) | Angle is between the two known sides |
| Two sides and a non‑included angle (SSA) | Law of Sines: a/sin(A) = b/sin(B) = c/sin(C) | May yield 0, 1, or 2 possible triangles (the ambiguous case) |
| Two angles and any side (ASA or AAS) | First find the third angle (sum = 180°), then apply the Law of Sines | Angle‑angle‑side or angle‑side‑angle |
| Right triangle with two legs known | Pythagorean Theorem: c² = a² + b² (c is hypotenuse) | One angle is exactly 90° |
| Right triangle with one leg and the hypotenuse known | Rearranged Pythagorean: a² = c² – b² | Same as above, solving for a leg |
Below we detail each method, provide the derivation where helpful, and walk through concrete examples.
1. Law of Cosines (SAS Case)
When you know two sides (a and b) and the angle (C) that lies between them, the third side (c) follows directly from the law of cosines:
[ c = \sqrt{a^{2} + b^{2} - 2ab\cos(C)} ]
Why it works: The formula extends the Pythagorean theorem to any triangle by subtracting a correction term that accounts for the angle not being 90°.
Example
Given: a = 7 cm, b = 10 cm, and the included angle C = 45°. Find c.
- Compute the cosine: cos 45° ≈ 0.7071.
- Plug into the formula:
[ c^{2} = 7^{2} + 10^{2} - 2(7)(10)(0.Now, 7071) \ c^{2} = 49 + 100 - 98. 99 \ c^{2} = 50 Not complicated — just consistent..
- Take the square root:
[ c ≈ \sqrt{50.01} ≈ 7.07\text{ cm} ]
A quick check: 7 cm + 10 cm > 7.07 cm and |10 − 7| < 7.07 cm, so the triangle inequality holds.
2. Law of Sines (SSA Case)
If you know two sides (a, b) and an angle (A) opposite one of those sides, the law of sines relates the ratios of side lengths to the sines of their opposite angles:
[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ]
To find the unknown side (c), you first need the angle opposite it (C). This often requires solving for the unknown angle using the known side‑angle pair, then using the angle sum property (A + B + C = 180°) to get the remaining angle.
Ambiguity note: With SSA, there may be zero, one, or two possible triangles. After computing a candidate angle via the inverse sine, check whether its supplement also yields a valid triangle.
Example
Given: a = 8 cm, b = 6 cm, and angle A = 30° (opposite side a). Find side c.
- Use the law of sines to find sin B:
[ \frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow \sin B = \frac{b\sin A}{a} ]
[ \sin B = \frac{6 \times \sin 30°}{8} = \frac{6 \times 0.5}{8} = \frac{3}{8} = 0.375 ]
- Determine angle B:
[ B = \arcsin(0.375) ≈ 22.0° ]
(The supplement 180° − 22° = 158° would make the sum of angles exceed 180° when added to A = 30°, so only the acute solution is valid.)
- Find the third angle C:
[ C = 180° - A - B = 180° - 30° - 22° = 128° ]
- Now apply the law of sines again to get c:
[ \frac{c}{\sin C} = \frac{a}{\sin A} \Rightarrow c = \frac{a\sin C}{\sin A} ]
[ c = \frac{8 \times \sin 128°}{\sin 30°} = \frac{8 \times 0.5} = \frac{6.That's why 7880}{0. 304}{0.5} = 12.
3. Using the altitude (height) and the Pythagorean theorem
When the triangle is not right‑angled, dropping a perpendicular from the vertex opposite the side of interest creates two right‑angled sub‑triangles. The altitude ( h ) can be expressed as
[ h = b\sin C = c\sin B, ]
and the projection of a side onto the base yields the remaining segment. By applying the Pythagorean theorem to either sub‑triangle we can isolate the unknown leg Most people skip this — try not to. That's the whole idea..
Example
Given a triangle with base (a = 14) cm, side (b = 9) cm, and the angle (C = 55^\circ) opposite side (c). Find the length of side (c) That's the part that actually makes a difference..
- Compute the altitude from the vertex opposite (a):
[ h = b\sin C = 9 \times \sin 55^\circ \approx 9 \times 0.8192 = 7.37\text{ cm}.
- Determine the projection of (b) onto the base (a):
[ p = b\cos C = 9 \times \cos 55^\circ \approx 9 \times 0.Practically speaking, 5736 = 5. 16\text{ cm} That alone is useful..
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The remaining part of the base is (a-p = 14 - 5.16 = 8.84) cm.
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Apply the Pythagorean theorem in the right triangle formed by the altitude, the projection, and side (c):
[ c = \sqrt{(a-p)^2 + h^2} = \sqrt{8.On top of that, 84^2 + 7. And 37^2} \approx \sqrt{78. 1 + 54.Because of that, 3} = \sqrt{132. On top of that, 4} \approx 11. 5\text{ cm}.
Thus side (c) is approximately 11.5 cm.
This approach is especially handy when a side and the angle adjacent to it are known, because the altitude and its projection follow directly from the sine and cosine of that angle Worth keeping that in mind..
4. Coordinate‑geometry method
Place the triangle in a Cartesian plane. To give you an idea, set one vertex at the origin and the opposite side along the x‑axis. If the known side lengths are (b) and (c) and the included angle is (\theta), the coordinates of the third vertex become ((b\cos\theta,; b\sin\theta)).
[ a = \sqrt{(c - b\cos\theta)^2 + (b\sin\theta)^2}. ]
Because the expression under the square root simplifies to (b^{2}+c^{2}-2bc\cos\theta), the coordinate method reproduces the law of cosines, offering a visual verification and a convenient way to handle problems where the triangle’s orientation forms part of the given information.
Example
Let (b = 6) cm, (c = 8) cm, and (\theta = 60^\circ). Compute (a):
[ \begin{aligned} a &= \sqrt{(8 - 6\cos 60^\circ)^2 + (6\sin 60^\circ)^2} \ &= \sqrt{(8 - 6 \times 0.Day to day, 0} \approx 7. 5)^2 + (6 \times \tfrac{\sqrt3}{2})^2} \ &= \sqrt{(8 - 3)^2 + (6 \times 0.Think about it: 0} \approx \sqrt{52. Even so, 196^2} \ &= \sqrt{25 + 27. 866)^2} \ &= \sqrt{5^2 + 5.21\text{ cm} Less friction, more output..
Conclusion
The law of cosines, the law of sines, the altitude‑projection technique, and the coordinate‑geometry approach each provide a reliable pathway to determine an unknown side — often called a “leg” — in a triangle. g.Day to day, , triangle inequality, angle sum), the desired leg can be solved efficiently and accurately. Also, the appropriate method depends on which measurements are given: side‑angle‑side calls for the cosine rule, side‑side‑angle (with careful handling of the ambiguous case) calls for the sine rule, while a known side together with an adjacent angle invites the use of altitude and right‑triangle relationships. Because of that, by selecting the method that matches the available data and verifying the triangle’s consistency (e. This versatility underscores the interconnectedness of the fundamental trigonometric tools introduced earlier and equips the reader with a flexible toolkit for any triangle‑solving scenario Most people skip this — try not to. Nothing fancy..
