Question

In the given figure ABCD is a parallelogram angle ADE equal to 50 degree and Angle ACE equals equals to angle BED equals to 90 degree the value of angle EAC + angle ABC - 2 angle DAC is

Answer 1

Hey buddy,



Given-

∠ADE = 50°

∠ACE = ∠BED = 90°


Solution-

Looking at the given figure,

∠ACD + ∠ACE = 180°

∠ACD = 180-90 = 90°   ….(∠ACE = 90°)

∠ADC + ∠ACD + ∠DAC = 180°  …( angles of triangle)

50 + 90 + ∠DAC = 180°

∠DAC = 180-50-90 = 40°            (1)

∠ABC = ∠ADC = 50°    (2) …(angles of parallelogram)

side AB parallel to side DE also ∠ACE = ∠BEC = 90°

Hence ABCE is a rectangle.


Diagonals of rectangle intersect each other at midpoint.

Thus triangle ABF becomes isolateral triangle.

∠EAB = ∠ABC = 50°

∠BAC = ∠BAE + ∠CAE

∠EAC = ∠BAC - ∠BAE = 90-50 = 40°    (3)


Putting vaules from (1), (2) & (3)

∠EAC + ∠ABC - 2∠DAC = 40 + 50 -2×40

∠EAC + ∠ABC - 2∠DAC = 90-80

∠EAC + ∠ABC - 2∠DAC = 10°


FINAL ANSWER

∠EAC + ∠ABC - 2∠DAC = 10°


Hope you got it...

Answer 2

Thank you for asking this question. Here is your answer:

We know that the Angle BED = 90°

And Angle AEC = 45°

Then the Angle EAB = 45°

Angle ABC is equal to 45°

Angle ADE = 50°

And Angle DAC = 40°

Angle EAC+ Angle ABC - 2 Angle DAC =  45° +45°-2×40°

= 90° - 80°

= 10°

So the final answer for this question is 10°

If there is any confusion please leave a comment below.