The angle C of an acute-angled triangle ABC, in which AB = √3, AC = 1, sin B = 1/2, is equal to …
| February 8, 2021 | education
We use the sine theorem for a triangle.
AC / SinABC = AB / SinACB = BC / SinBAC.
Then SinACB = AB * SinABC / AC = √3 * (1/2) / 1 = √3 / 2.
Angle ACB = arcsin (√3 / 2) = 600.
Answer: Angle ACB is 600.
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