M∠ACB = 52° , find m∠CBE =°

(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=6

√

2

 cm

Thus, d(B, C) = 6

√

2

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.

Step-by-step explanation:

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.