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Question 24 A parallelogram has sides of lengths 10 and 6, and one angle is 42°. The length of the longer diagonal is: $(36 + 100 - 120 \cos 138^\circ)^{\frac{1}{2}}$ $(36 + 100 - 120 \cos 42^\circ)^{\frac{1}{2}}$ $6 + 10 - \sqrt{120 \cos 42^\circ}$ None of these expressions I don't know.

          Question 24
A parallelogram has sides of lengths 10 and 6, and one angle is 42°. The length of the longer diagonal is:
$(36 + 100 - 120 \cos 138^\circ)^{\frac{1}{2}}$
$(36 + 100 - 120 \cos 42^\circ)^{\frac{1}{2}}$
$6 + 10 - \sqrt{120 \cos 42^\circ}$
None of these expressions
I don't know.

Question 24
A parallelogram has sides of lengths 10 and 6, and one angle is 42°. The length of the longer diagonal is:
(36 + 100 - 120 cos 138^∘)^(1)/(2)
(36 + 100 - 120 cos 42^∘)^(1)/(2)
6 + 10 - √(120 cos 42^∘)
None of these expressions
I don't know.

Added by Heather E.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Donna Densmore

07/15/2023

Step 1

The law of cosines states that for a triangle with sides a, b, and c, and angle C opposite side c, the following equation holds: c^2 = a^2 + b^2 - 2ab*cos(C) Show more…

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Question 24 A parallelogram has sides of lengths 10 and 6, and one angle is 42°. The length of the longer diagonal is: (36 + 100 - 120cos138°)^(1/2) (36 + 100 - 120cos42°)^(1/2) 6 + 10 - √(120cos42°) None of these expressions I don't know.