Mathematics, 18.12.2020 03:50, toot3715
Given square JKLM (T(-6,4)oT(1,5)) (JLKM)=RSTU, what is the area of RSTU?
Answers: 2
answered: Guest
answer: cos [ arcsin (3/5) - arccos (1/2) ] =
step-by-step explanation: Ă— = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg
cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13
=
0.12
b. cos (36.87 - 60) = cos (-23.23) = 0.92
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Given square JKLM (T(-6,4)oT(1,5)) (JLKM)=RSTU, what is the area of RSTU?...
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