Instruction

1

To calculate the percentage of any of the angles of an arbitrary

**triangle**use the theorem of cosines. It States that the square**of the length of**any*of the parties*s (e.g., A) equal to the sum of the squares of the lengths of the other two*sides*(B and C), from which is subtracted the product of their lengths, the cosine of the angle (α) lying in formed their top. This means that you can Express the cosine through**the length of the***sides*: cos(α) = (B2+C2-A2)/(2*A*B). To get the value of this angle in degrees, to the expression, apply the inverse cosine function the inverse cosine: α = arccos((B2+C2-A2)/(2*A*B)). This way you calculate the value of one of the corners - in this case, which lies opposite the*sides*s of A.2

To calculate the two remaining corners, you can use the same formula, changing it in some places

**lengths**of the known*sides*. But a simpler expression with less number of mathematical operations can be obtained utilizing another postulate from the field of trigonometry the theorem of sines. She claims that the ratio of**the lengths of**any*of the parties*s to the sine of the opposite angle in her triangle are equal. This means that you can Express, for example, the sine of the angle β, which lies opposite*parties*s B across the length of the*sides*s C and the already calculated angle α. Multiply the length of B sine of α, and the result divide by the length of the C sin(β) = B*sin(α)/C. the Value of this angle in degrees, as in the previous step, calculate using inverse trigonometric functions - this time of the arcsine: β = arcsin(B*sin(α)/C).3

The magnitude of the remaining angle (γ) can be calculated for any of the obtained in the previous steps, formulas, changing them places

**the length of the***sides*. But it's easier to use another theorem about the sum of the angles in the triangle. She argues that this sum is always 180°. Since two of the three angles you have already known, just subtract from 180° the value to get the value of the third: γ = 180°-α-β.