Trigonometry?

Question

in this:
http://tinyurl.com/2w3pxg
CD=5
BC=6
BE=3
angle D=angle C=90 degrees
calculate angle EOC

the picture is in the link

I said that angle D = Angle C they are both 90 degrees

Answer 1

Assume O is the intersection point of AE and CD.
BD = √(9^2-5^2) = √56
CD^2 = AD*BD, geometric mean
25 = AD*√56
AD = 25/√56

Once you know AD, you can easily find AC, angle AEC and angle ECO.

angle EOC = 180 - angle AEC - angle ECO = 60.26 deg, the final answer.

Can you finish it now?

Answer 2

If BC = 6
BE = 3
then, EC = 3

CD = 5

cos BCD = 5/6
cos BCD = 0.8333
Angle BCD = 34º

Assuming the point O is between the intersection
of the point A and E, and between point D and C

Angle BCD = Angle ECO

tan Angle ECO = EO/3
ECO = 34º

OE = (tan 34º) (3)
OE = 2

tan Angle OEC = 3/2
tan Angle OEC = 1.5 perform arc tan, will be
tan Angle OEC = 56.30º

angle OEC + angle ECO = 90.3

180º is the sum of all angles in a triangle,

EOC = 180 - 90.3
= 89.7 say, 90º answer.

Answer 3

I assume O is the intersection of CD and AE
BCD is a right triangle so BD= sqrt(36-25) = sqrt(11).ABC is a right triangle with altitude CD, so CD^2= BD*DA.
This gives DA = 25/sqrt(11). Since ADC is a right triangle AC can be computed as sqrt(AD^2)CD^2.
Since EC = 3

Answer 4

where's O on the picture?
and what are you saying here: angle D=angle