☰ Contents · Geometry

The law of sines

Lessons 28 · 1 lessons · B. Xaydarov, E. Sariqov, A. Qo‘chqorov. Geometry, Grade 9 (textbook for general secondary schools), revised 4th edition. Huquq va Jamiyat Publishing, Tashkent, 2019
28

The law of sines

Textbook: pp. 84–85
GoalKnow the law of sines and use it to find a side or an angle and the radius of the circumscribed circle.
New words
law of sines · sinuslar teoremasicircumscribed circle · tashqi chizilgan aylanaopposite angle · qarshisidagi burchakcorresponding ratio · mos nisbat
Explanation

The law of sines: the sides of a triangle are proportional to the sines of the opposite angles: a/sinA = b/sinB = c/sinC. It follows by equating the three expressions of the area, S = ½ab·sinC = ½bc·sinA = ½ac·sinB. This common ratio equals the diameter of the circumscribed circle: a/sinA = 2R. With it we find the other sides when two angles and a side are given, and the sine of another angle when two sides and an angle opposite one of them are given. Note: the equation sinB = x may have two solutions between 0° and 180° (B and 180° − B), so the existence of the triangle must be checked.

Worked examples
△ABC: ∠A = 30°, ∠B = 45°, a = BC = 8. b = a·sinB/sinA = 8·(√2/2)/(1/2) = 8√2.
a = 10, ∠A = 30°: 2R = a/sinA = 10/(1/2) = 20, so R = 10.
Class activity

“Opposite angle”: the teacher draws a triangle and gives two sides and some of the opposite angles; pairs use the law of sines to find the missing ones.

Practice
1
a = 6, ∠A = 30°, ∠B = 90°. Find b.
2
∠A = 45°, ∠B = 30°, a = 10√2. Find b.
3
a = 9 and ∠A = 60°. What is the radius of the circumscribed circle?
4
Why can there be two triangles when a = 5, ∠A = 30° and b = 5√3?