Question

Given: ABCD trapezoid, BK ⊥ AD , AB=DC AB=8, AK=4 Find: m∠A, m∠B

Answer 1

Answer:

$m\angle B=m\angle C=120^{\circ}$

$m\angle A=m\angle D=60^{\circ}$

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

• $\angle A\cong \angle D;$
• $\angle B\cong \angle C.$

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle,  AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

$AB=2AK.$

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

$m\angle ABK=30^{\circ}$

Since BK ⊥ AD, then BK ⊥ BC and

$m\angle KBC=90^{\circ}$

Thus,

$m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}$

Now,

$m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}$