Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
1/2 = cos(2*t)sqrt(3) /2 = sin(-2*t)find tfind the time derivative of x(t), y(t)(-2sin(2*t) , -2cos(2*t) )plug in the t you found to find the velocity vector.
Answer:
The answer is 10x^6 * sqrt x
Step-by-step explanation:
Hope it works
