What is the length of side b is a question that appears frequently in geometry, trigonometry, and real‑world problem solving. Whether you are working with a right triangle, an oblique triangle, or a coordinate‑geometry figure, determining the measure of side b often requires applying a specific formula based on the information you already know. This article walks you through the concepts, methods, and step‑by‑step calculations you need to confidently find side b in various contexts, while highlighting common pitfalls and offering practice examples Nothing fancy..
Introduction
In any triangle, sides are conventionally labeled a, b, and c, with each side opposite its corresponding angle A, B, and C. When a problem asks what is the length of side b, it is seeking the numerical value of the side that lies opposite angle B. The approach you take depends on which other measurements are given: side lengths, angles, or a combination of both. Below we outline the most reliable techniques and illustrate each with clear examples.
Understanding Side b in Different Triangle Types
1. Right Triangles
If the triangle contains a 90° angle, side b may be one of the legs or the hypotenuse. The Pythagorean theorem is the go‑to tool:
[ a^{2} + b^{2} = c^{2} ]
where c is the hypotenuse. Knowing any two sides lets you solve for the third.
2. Oblique (Non‑Right) Triangles
When no angle equals 90°, you must rely on the Law of Sines or the Law of Cosines. These laws relate side lengths to the sines and cosines of their opposite angles Worth knowing..
- Law of Sines
[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ]
- Law of Cosines
[ b^{2} = a^{2} + c^{2} - 2ac\cos B ]
Choosing between them hinges on what data you have:
- Use the Law of Sines when you know one side and its opposite angle, plus another angle or side.
- Use the Law of Cosines when you know two sides and the included angle, or when you know all three sides and need to find an angle (then rearrange to solve for a side).
Methods to Find the Length of Side b
Below is a concise decision‑tree you can follow:
| Known Information | Recommended Formula | Steps |
|---|---|---|
| Two sides and the included angle (SAS) | Law of Cosines | Plug into (b^{2}=a^{2}+c^{2}-2ac\cos B) and solve for b. |
| One side, its opposite angle, and another angle (ASA or AAS) | Law of Sines | Use (\frac{b}{\sin B} = \frac{known;side}{\sin;known;angle}) → solve for b. |
| Two sides and a non‑included angle (SSA) – ambiguous case | Law of Sines (check for one/two solutions) | Compute possible b values; verify with triangle inequality. |
| Right triangle, legs known | Pythagorean theorem | If b is a leg: (b = \sqrt{c^{2}-a^{2}}). If b is the hypotenuse: (b = \sqrt{a^{2}+c^{2}}). |
| Right triangle, one leg and an angle known | Trigonometric ratios (sine, cosine, tangent) | Example: (b = a \tan B) if a is adjacent to angle B. |
Step‑by‑Step Examples
Example 1: Using the Law of Cosines (SAS)
Problem: In triangle ABC, side a = 7 cm, side c = 9 cm, and the included angle B = 48°. Find the length of side b The details matter here. That alone is useful..
Solution:
- Write the Law of Cosines for side b:
[ b^{2} = a^{2} + c^{2} - 2ac\cos B ]
- Substitute the known values:
[ b^{2} = 7^{2} + 9^{2} - 2(7)(9)\cos 48^{\circ} ]
- Compute the squares and product:
[ b^{2} = 49 + 81 - 126\cos 48^{\circ} ]
- Evaluate (\cos 48^{\circ}) (≈ 0.6691):
[ b^{2} = 130 - 126(0.6691) = 130 - 84.30 = 45.
- Take the square root:
[ b = \sqrt{45.70} \approx 6.76\text{ cm} ]
Answer: Side b ≈ 6.8 cm (rounded to one decimal place) And that's really what it comes down to..
Example 2: Using the Law of Sines (AAS)
Problem: In triangle DEF, angle D = 35°, angle E = 70°, and side d (opposite angle D) = 12 in. Find side e (opposite angle E), which we will label as side b for consistency Less friction, more output..
Solution:
- First find the third angle:
[ F = 180^{\circ} - D - E = 180^{\circ} - 35^{\circ} - 70^{\circ} = 75^{\circ} ]
- Apply the Law of Sines:
[ \frac{d}{\sin D} = \frac{e}{\sin E} ]
- Insert known values:
[ \frac{12}{\sin 35^{\circ}} = \frac{e}{\sin 70^{\circ}} ]
- Solve for e:
[ e = \frac{12 \sin 70^{\circ}}{\sin 35^{\circ}} ]
- Compute the sines (sin 35° ≈ 0.574, sin 70° ≈ 0.940):
[ e = \frac{12 \times 0.940}{0.And 574} \approx \frac{11. Here's the thing — 28}{0. 574} \approx 19.
Answer: Side b (here e) ≈ 19.7 in.
