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TrianglesGeometryHigh schoolabout 1:11

Solve the triangle with sides a = 10, b = 10, c = 12: animated lesson

A narrated, step-by-step triangles video lesson with captions. Final answer: a = 10, b = 10, c = 12; A = 53.13010235°, B = 53.13010235°, C = 73.73979529°; area = 48, perimeter = 32.

SpeedLanguage

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.4 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
Solve the triangle with sides a = 10, b = 10, c = 12.
2. Step by step
1. $\cos A = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow A \approx 53.13010235^\circ,\ B \approx 53.13010235^\circ,\ C \approx 73.73979529^\circ$
  Each angle from the law of cosines
2. $\text{Area} = \tfrac12ab\sin C \approx 48,\ P \approx 32$
  Area and perimeter
3. The answer
$\text{Area} \approx 48$
area ≈ 48
4. Check
- Triangle 1: angles sum to 180°; a/sin A = 12.5 = c/sin C = 12.5.

Transcript

The problem. Here is the problem: Solve the triangle with sides a equals 10, b equals 10, c equals 12. Side a is opposite angle A, b opposite B and c opposite C. This is the SSS case.

Step by step. Each angle from the law of cosines. Area and perimeter.

The answer. So the answer is a equals 10, b equals 10, c equals 12; A equals 53.13010235°, B equals 53.13010235°, C equals 73.73979529°; area equals 48, perimeter equals 32. Three measurements including a side normally fix a triangle. Two sides with a non-included angle (SSA) can give zero, one or two triangles.

Check. Triangle 1: angles sum to 180°; a/sin A equals 12.5 equals c/sin C equals 12.5.

About triangles

Solve triangles with the law of sines and the law of cosines, and find their area.

Solve the triangle with sides a = 10, b = 10, c = 12