Given: klmn is a trapezoid, kl=mn, ab is a midsegment, a f=2, fb=5, mn=6 find: m∠n, m∠k, m∠l, and m∠m

∠K = ∠N = 60°

∠L = ∠M = 120°

In the attached, we have renamed F to G so will let us talk about it more easily. We have also added altitude MX.

AG is also a midsegment of ΔKLM, so LM = 2×AG = 4. Then ...

... NX = AB - LM = 7 -4 = 3

and we have right ΔMXN with hypotenuse 6 and leg 3. This is recognizable as a 30°-60°-90° triangle, with the 60° angle at N.

The angle at M is supplementary to that at N (because LM ║ KN), so measures 120°

The trapezoid is isosceles, so angles K and L have the same measures as angles N and M.

answer: b.48 for

step-by-step explanation: