Question

The diagonal of a rectangle ABCD meet at K. If Angle AKB = 140° Find Angle ACB ?


Please answer it with a solution. THANK YOU

Answer 1

∠ACB = 70°

Step-by-Step Explaination:

Given,ABCD is a rectangle.

∠AKB =140°

∠AKB  = ∠CKD (Vertically Opposite Angle)  

∠CKD = 140°

Also, DK = CK (Since diagonal of a rectangle are equal and bisect each other)

∠KDC = ∠KCD (Angle opposite to equal sides are always equal)

Now,In Δ KCD

∠KDC  + ∠KCD + ∠DKC = 180° (Angle sum property of a triangle)

∠KDC + ∠KDC + ∠DKC = 180°

2∠KDC + 140° = 180°

2∠KDC = 180°-140°

∠$KDC = \frac{40}{2}$°

∠KDC = 20°

⇒∠KCD = 20° = ∠ACD

∠BCD = 90° (All the angles are right angle in a rectangle)

∠BCD = ∠BCA + ∠ACD

90° =  ∠BCA + 20°

∠BCA = 90° - 20°

∠BCA = 70°

or ∠ACB = 70°