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∠TVX = 49°

x = 52°

x = 57.5°

BC = 6 cm

Step-by-step explanation:

Question 1

Angle at the Centre Theorem:

∠UTV = 1/2 ∠UOV = 134 ÷ 2 = 67°

Angles around a point add up to 360°

⇒ interior angle of ∠UOV = 360 - 134 = 226°

The sum of interior angles of a non-cyclic quadrilateral is 360°

⇒ ∠OVT = 360 - 26 - 226 - 67 = 41°

As WVX is a tangent, it forms a right angle with the radius (OV)

⇒ ∠TVX = 90 - 41 = 49°

Question 2

Angles on a straight line add up to 180°

The angle between a tangent and a chord is equal to the angle in the alternate segment.

Therefore ∠x = 180 - 68 - 60 = 52°

Question 3

As CD is a tangent ⇒ OC = radius ⇒ ∠OCD = 90°

The sum of interior angles of a triangle is 180°

⇒ ∠COD = 180 - 90 - 25 = 65°

As ∠COD =∠AOB then ∠AOB = 65°

As OB and OA are radii, triangle AOB is an isosceles triangle. Therefore ∠OBA = ∠OAB

Therefore x = (180 - 65) ÷ 2 = 57.5°

Question 4

As BC is the tangent to the smaller circle, OA is perpendicular to BC.

Therefore, triangle OAB is a right triangle with hypotenuse 5 cm.

Using Pythagoras' Theorem a² + b² = c² (where a and b are the legs, and c is the hypotenuse of a right triangle):

⇒ AB² + 4² = 5²

⇒ AB = √(5² - 4²) = 3

⇒ BC = 2 x AB = 3 x 2 = 6 cm