Find the longer diagonal of a parallelogram having sides of lengths 10 and 15 and an angle measure 60. (round to the nearest whole number.)

You can use the "Law of cosines":

c^2 = a^2 + b^2 - 2 a b cos C

where a, b and c are the three sides of a triangle, and C is the opposite angle to side c.

In this case, the triangle is half a parallelogram. Sides a and b of the triangle are the sides of the parallelogram, and side c of the triangle is the longest diagonal of the parallelogram. So:

c = x; a = 10; b =15

The opposite angle to the diagonal is the biggest angle of the parallelogram, because we want the longest diagonal. The angles in a parallelogram are supplementary. Because the given angle is acute, it is the small one. Hence, angle C equals 180 - 60 = 120.

Plugging all this into the "Law of cosines", you get:

x^2 = 10^2 + 15^2 - 2·10·15·cos 120 = 175

x = √175 = 13.22

Rounded to the nearest whole number:

x ≈ 13

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