Answer:
The solution is shown below
Explanation:
From the image attached:
Line BC and line AC are the radii of the circle, therefore BC = AC.
Since BC = AC, the triangle ABC is an isosceles triangle with m∠CBA = m∠CAB = 28°
Also line ED is a tangent to circle C. From the circle theorems, a tangent to a circle is at right angle to the circle radius at point of contact. Since AC is a radii, ∠EAC = 90°
Therefore:
∠DAB + ∠EAC + m∠CAB = 180° (sum of angles on a straight line)
∠DAB = 180° - ∠EAC - m∠CAB
∠DAB = 180 - 90 - 28 = 62
∠DAB = 62°
