How to Calculate the Sides in a Triangle?

Triangles are one of the most fundamental shapes in geometry, and there are several methods to calculate their sides depending on the type of triangle and the information available. Let’s explore these methods in detail.

1. Using the Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, primarily used for right-angled triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

Formula

The Pythagorean theorem is expressed as:

$c^2 = a^2 + b^2$

Where:

  • $c$ is the length of the hypotenuse.
  • $a$ and $b$ are the lengths of the other two sides.

Example

Suppose you have a right-angled triangle with one side of 3 units and another side of 4 units. To find the hypotenuse:

$c^2 = 3^2 + 4^2$

$c^2 = 9 + 16$

$c^2 = 25$

$c = text{sqrt}(25)$

$c = 5$

So, the hypotenuse is 5 units long.

2. Using Trigonometric Ratios

Trigonometric ratios are useful for calculating the sides of a triangle when you know one angle and one side. The primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).

Formulas

  • Sine: $text{sin}(theta) = frac{text{opposite}}{text{hypotenuse}}$
  • Cosine: $text{cos}(theta) = frac{text{adjacent}}{text{hypotenuse}}$
  • Tangent: $text{tan}(theta) = frac{text{opposite}}{text{adjacent}}$

Example

Suppose you have a right-angled triangle with an angle of 30 degrees and a hypotenuse of 10 units. To find the length of the side opposite the angle:

$text{sin}(30^text{°}) = frac{text{opposite}}{10}$

$0.5 = frac{text{opposite}}{10}$

$text{opposite} = 0.5 times 10$

$text{opposite} = 5$

So, the length of the side opposite the 30-degree angle is 5 units.

3. Using the Law of Sines

The Law of Sines is useful for any triangle, not just right-angled ones. It relates the lengths of the sides of a triangle to the sines of its angles.

Formula

$frac{a}{text{sin}(A)} = frac{b}{text{sin}(B)} = frac{c}{text{sin}(C)}$

Where:

  • $a$, $b$, and $c$ are the lengths of the sides.
  • $A$, $B$, and $C$ are the angles opposite these sides.

Example

Suppose you have a triangle with sides $a = 7$, $b = 9$, and angle $A = 30^text{°}$. To find the angle $B$:

$frac{7}{text{sin}(30^text{°})} = frac{9}{text{sin}(B)}$

$frac{7}{0.5} = frac{9}{text{sin}(B)}$

$14 = frac{9}{text{sin}(B)}$

$text{sin}(B) = frac{9}{14}$

$B = text{sin}^{-1}bigg(frac{9}{14}bigg)$

$B text{≈ 41.81 degrees}$

4. Using the Law of Cosines

The Law of Cosines is another useful tool for any triangle. It is especially handy when you know two sides and the included angle or all three sides.

Formula

$c^2 = a^2 + b^2 – 2ab text{cos}(C)$

Where:

  • $a$, $b$, and $c$ are the lengths of the sides.
  • $C$ is the included angle.

Example

Suppose you have a triangle with sides $a = 5$, $b = 6$, and angle $C = 60^text{°}$. To find the length of side $c$:

$c^2 = 5^2 + 6^2 – 2 times 5 times 6 times text{cos}(60^text{°})$

$c^2 = 25 + 36 – 60 times 0.5$

$c^2 = 25 + 36 – 30$

$c^2 = 31$

$c = text{sqrt}(31)$

$c text{≈ 5.57 units}$

Conclusion

Understanding how to calculate the sides of a triangle is essential in geometry. Whether you use the Pythagorean theorem, trigonometric ratios, the Law of Sines, or the Law of Cosines, each method provides a unique way to solve for unknown sides. Knowing these techniques will help you tackle a wide range of problems, from simple right-angled triangles to more complex non-right-angled ones.

3. Wikipedia – Law of Sines4. Wikipedia – Law of Cosines

Citations

  1. 1. Khan Academy – Calculating Triangle Sides
  2. 2. Math is Fun – Trigonometry

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