Thread: Angles... :(

In the figure, ABCD is a square and ABE is an equilateral triangle. Find <ADE. I just for the life of me find out what E and D are... I got A, but... Grrrr...

And then another question... I don't get where my teacher gets this stuff...

(x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4) is an identity in x, find the values of A and B.

Any and all help is appreciated!

Originally Posted by Rocher

In the figure, ABCD is a square and ABE is an equilateral triangle. Find <ADE. I just for the life of me find out what E and D are... I got A, but... Grrrr...

AB = AE because triangle AEB is equilateral.

AD = AB because ABCD is a square.

AD = AE because AD = AB = AE.

Thus triangle EAD is isosceles.

Thus angle ADE = angle AED.

Angle DAE is the complement of angle EAB because the interior angles of a square are right angles.

Angle EAB is 60 degrees because triangle AEB is equilateral.

Thus angle DAE is 90 - 60 = 30 degrees.

Call angle AED x. Then 30 + x + x = 180 because the sum of the interior angles of a triangle is 180 degrees.

Solving this equation I get that x = 75, so angle AED = angle ADE= 75 degrees.

-Dan

Originally Posted by Rocher

And then another question... I don't get where my teacher gets this stuff...

(x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4) is an identity in x, find the values of A and B.

Any and all help is appreciated!

This really should have gone in its own thread. See rule #15 here.



Expanding


Collecting like terms:




In order for this expression to be 0 for all x we must have the coefficients of all powers of x equal to 0. So

and


Can you take things from here?

-Dan