Given: abcd - rectangle

area of abcd = 458m2

m∠aob = 80°

find: ab, bc

AB = 19.60 m and BC = 23.37 m to the nearest hundredth.

Step-by-step explanation:

The sides OA OB OC and OD are all equal and we have 2 pairs of equal isosceles triangles formed.

One with angles 80 & 2*50 and another pair with angles 100 and 2*40 degrees.

Consider the triangle with AB as the base.

Applying the Sine Rule we have:

AB / sin 80  = OA  / sin 50  so  AB = OA sin 80 / sin 50

Consider now the triangle BOC:

BC / sin 100 = OB / sin 50 = OA / sin 50 ( because OA = OB).

So BC = OA sin 100 / sin 40

Therefore AB /BC =   (OA sin 80 / sin 50)  /  (OA sin 100 / sin 40)

=  sin 80 * sin 40 / sin 50 sin 100

= 0.8391

and BC = AB / 0.8391.

Now we are given that the area = AB * BC = 458.

Therefore substituting for BC we have:

AB * AB / 0.8391 = 458

AB^2 = 458 * 0.8391

=  384.31

AB = 19.60

and BC = 458/19.60

= 23.37.