Asked by joe
                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
                    Answered by
            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.
                    Answered by
            joanne
            
    wow this really helps!! :D I have the same problem too haha thanks!!!
    
                    Answered by
            Paul
            
    Hey thanks! I have the same problem. never thought it would be this easy. 
    
                    Answered by
            daryl
            
    tnx bro!
    
                    Answered by
            balh blahw
            
    where'd you get that 10 and 12 
    
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