∠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
