i. line AB is the tangent to the circle with centre C and radius AC. [Given]
∴ ∠CAB = 90° (i) [Tangent theorem]
ii. seg CA ⊥ line AB [From (i)]
radius = l(AC) = 6 cm
∴ The distance of point C from line AB is 6 cm.
iii. In ∆CAB, ∠CAB = 90° [From (i)]
∴ BC2 = AB2 + AC2 . [Pythagoras theorem]
= 62 + 62
= 2 × 62
∴ BC = \(\sqrt{2 \times 6^{2}}\) [Taking square root of both sides]
= 6 \(\sqrt { 2 }\) cm
∴ d(B, C) = 6 cm
iv. In ∆ABC,
AC = AB = 6cm
∴ ∠ABC = ∠ACB [Isosceles triangle theorem]
Let ∠ABC = ∠ACB =x
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ 90° + x + x = 180°
∴ 90 + 2x = 180°
∴ 2x = 180°- 90°
∴ x = \(\frac{90^{\circ}}{2}\)
∴ x = 45°
∴ ∠ABC = 45°