How to Find the Length of a Triangle Side?

Finding the length of a side in a triangle can be approached in several ways, depending on the type of triangle and the information available. Let’s explore some methods.

Using the Pythagorean Theorem

If you have a right-angled triangle, the Pythagorean Theorem is a straightforward way to find the length of a side. The theorem states:
$a^2 + b^2 = c^2$
where $c$ is the hypotenuse (the side opposite the right angle), and $a$ and $b$ are the other two sides. For example, if $a = 3$ and $b = 4$, then $c$ can be found as follows:
$c = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$

Using the Law of Sines

The Law of Sines is useful for any triangle, not just right-angled ones. It states:
$frac{a}{sin(A)} = frac{b}{sin(B)} = frac{c}{sin(C)}$
where $a$, $b$, and $c$ are the sides of the triangle, and $A$, $B$, and $C$ are the opposite angles. For instance, if you know $a = 7$, $A = 30degree$, and $B = 45degree$, you can find $b$ as follows:
$frac{7}{sin(30degree)} = frac{b}{sin(45degree)}$
$frac{7}{0.5} = frac{b}{frac{sqrt{2}}{2}}$
$14 = b cdot frac{sqrt{2}}{2}$
$b = 14 cdot frac{2}{sqrt{2}} = 14 cdot sqrt{2}$

Using the Law of Cosines

The Law of Cosines is another powerful tool, particularly when you know two sides and the included angle. It states:
$c^2 = a^2 + b^2 – 2ab cdot cos(C)$
For example, if $a = 5$, $b = 6$, and $C = 60degree$, you can find $c$ as follows:
$c^2 = 5^2 + 6^2 – 2 cdot 5 cdot 6 cdot cos(60degree)$
$c^2 = 25 + 36 – 60 cdot 0.5$
$c^2 = 25 + 36 – 30$
$c^2 = 31$
$c = sqrt{31}$

Using the Triangle Inequality Theorem

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This can help confirm if the given sides can form a triangle. For example, if you have sides of lengths 3, 4, and 8, check:
$3 + 4 > 8$ (False)
$3 + 8 > 4$ (True)
$4 + 8 > 3$ (True)
Since one inequality is false, these sides cannot form a triangle.

Conclusion

Understanding these methods helps solve various triangle problems. Whether using the Pythagorean Theorem for right-angled triangles, the Law of Sines and Cosines for any triangle, or the Triangle Inequality Theorem for validation, you have the tools to find the length of a triangle side.

Citations

  1. 1. Khan Academy – Pythagorean Theorem
  2. 2. Math is Fun – Law of Sines
  3. 3. Math is Fun – Law of Cosines

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ