The sum of the two sides of the triangle is 16 cm, and the angle between them is 120 °.

| April 25, 2021 | education

The sum of the two sides of the triangle is 16 cm, and the angle between them is 120 °. Find the smaller of these sides if the third side of the triangle is 14 cm.

Let a triangle ABC be given, with sides AB, BC, AC, angle <B = 120 °, AB + BC = 16 cm, AC = 14 cm.
Then AB = 16-BC.
According to the cosine theorem:
AC² = AB² + BC²-2AB * BC * cosB, substituting the numbers we have:
14² = AB² + BC²-2AB * BC * (- 1/2).
169 = AB² + BC² + AB * BC.
Substitute AB into this expression:
169 = AB² + BC² + AB * BC = (16-BC) ² + BC² + (16-BC) * BC.
169 = 256-32 * ВС + ВС2 + ВС2 + 16 * ВС-ВС2.
ВС²-16 * ВС + 60 = 0.
Let’s solve the quadratic equation:
D = (- 16) ²-4 * 1 * 60 = 16.
x1 = 16-√16 / 2 * 1 = 16-4 / 2 = 6 cm.
x2 = 16 + √16 / 2 * 1 = 16 + 4/2 = 10 cm.
Both roots satisfy us, then we substitute both options and find the missing side:
AB = 16-BC = 16-6 = 10 cm, or AB = 16-BC = 16-10 = 6 cm.
The remaining sides of the triangle are 6 and 10 cm.
Answer: the smaller side is 6 cm.



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