If N is a point on the straight line AB and PN is perpendicular to AB , then prove that
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(1) It is given that line AB is tangent to the circle at A.
∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)
Thus, the measure of ∠CAB is 90º.
(2) Distance of point C from AB = 6 cm (Radius of the circle)
(3) ∆ABC is a right triangle.
CA = 6 cm and AB = 6 cm
Using Pythagoras theorem, we have
BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cm
Thus, d(B, C) = 62–√ cm
(4) In right ∆ABC, AB = CA = 6 cm
∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)
Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)
∴ 2∠ABC = 90º
⇒ ∠ABC = 90°2 = 45º
Thus, the measure of ∠ABC is 45º.
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