How to Solve a Spherical Triangle?

Solving a spherical triangle is a fascinating topic in spherical trigonometry, which deals with triangles on the surface of a sphere. Unlike flat triangles, spherical triangles have curved sides, making the math a bit more complex. Let’s break down the steps to solve a spherical triangle.

Key Concepts

Spherical Triangle

A spherical triangle is formed by three great circle arcs, which are the intersection of the sphere with planes passing through its center. The angles of a spherical triangle are always greater than the corresponding angles of a planar triangle.

Notation

In spherical trigonometry, the sides of the triangle are usually denoted by lowercase letters (a, b, c), and the angles opposite these sides are denoted by uppercase letters (A, B, C).

Formulas

To solve a spherical triangle, you can use the spherical law of cosines and the spherical law of sines.

Spherical Law of Cosines

For sides:

$cos a = cos b cos c + sin b sin c cos A$

For angles:

$cos A = -cos B cos C + sin B sin C cos a$

Spherical Law of Sines

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

Example Problem

Suppose we have a spherical triangle with sides $a = 60^text{°}$, $b = 70^text{°}$, and angle $C = 80^text{°}$. We want to find angle $A$

Step-by-Step Solution

  1. Use the spherical law of cosines for angles:

$cos A = -cos B cos C + sin B sin C cos a$

  1. Plug in the known values:

Since we don’t know $B$ yet, we can use another approach. We can use the spherical law of cosines for sides first to find side $c$:

$cos c = cos a cos b + sin a sin b cos C$

  1. Calculate $cos c$:

$cos c = cos 60^text{°} cos 70^text{°} + sin 60^text{°} sin 70^text{°} cos 80^text{°}$

  1. Simplify the expression:

$cos c = 0.5 times 0.3420 + 0.8660 times 0.9397 times 0.1736$

  1. Find $c$:

$cos c = 0.1710 + 0.1405 approx 0.3115$

$c approx cos^{-1}(0.3115) approx 71.8^text{°}$

  1. Now use the spherical law of cosines for angles:

$cos A = frac{cos a – cos b cos c}{sin b sin c}$

  1. Plug in the known values:

$cos A = frac{cos 60^text{°} – cos 70^text{°} cos 71.8^text{°}}{sin 70^text{°} sin 71.8^text{°}}$

  1. Simplify the expression:

$cos A = frac{0.5 – 0.3420 times 0.3115}{0.9397 times 0.9461}$

  1. Calculate $A$:

$cos A = frac{0.5 – 0.1065}{0.8890} approx 0.442$

$A approx cos^{-1}(0.442) approx 63.8^text{°}$

So, angle $A$ is approximately $63.8^text{°}$

Conclusion

Solving a spherical triangle may seem daunting at first, but with the right formulas and a step-by-step approach, it becomes manageable. These principles are crucial in fields like astronomy and navigation, where understanding the geometry of the sphere is essential.

3. Britannica – Spherical Trigonometry

Citations

  1. 1. Khan Academy – Spherical Trigonometry
  2. 2. Wolfram MathWorld – Spherical 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