Question
the bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65 degree and 40 degree respectively find the two legs. Answer using law of sines pls
Answers
Reiny
Construct the trapezoid ABCD ,where AB || CD
AB = 12 and CD = 22
angle C=65 and angle D = 40
Draw AE || BC where E is on CD
So now ABCE is a parallelogram, and CE = 12
which makes ED = 10
Now look at triangle AED, by corresponding angles
angle AED = 65°, angle D = 40 leaving angle DAE = 75°
by sine law:
AD/sin65 = 10/sin75
AD = 10sin65/sin75 = 9.38
by sine law:
AE/sin40 = 10/sin75
AE = 6.65
but BC = AE, (||gram|
So the side adjacent to the 65° angle is 6.65, the side adjacent to the 40° angle is 9.38
check my arithmetic, I am only on my first coffee.
AB = 12 and CD = 22
angle C=65 and angle D = 40
Draw AE || BC where E is on CD
So now ABCE is a parallelogram, and CE = 12
which makes ED = 10
Now look at triangle AED, by corresponding angles
angle AED = 65°, angle D = 40 leaving angle DAE = 75°
by sine law:
AD/sin65 = 10/sin75
AD = 10sin65/sin75 = 9.38
by sine law:
AE/sin40 = 10/sin75
AE = 6.65
but BC = AE, (||gram|
So the side adjacent to the 65° angle is 6.65, the side adjacent to the 40° angle is 9.38
check my arithmetic, I am only on my first coffee.
joanne
wow this really helps!! :D I have the same problem too haha thanks!!!
Paul
Hey thanks! I have the same problem. never thought it would be this easy.
daryl
tnx bro!
balh blahw
where'd you get that 10 and 12