Answer:
(a) 52°
(b) 322°
Step-by-step explanation:
(a) The details of the circle are;
The diameter of the circle = AOC
The center of the circle = Point O
The point the line AT cuts the circle = Point B
The point the tangent PT touches the circle = Point C
Angle ∠COB = 76°
We have that angle AOB and angle COB are supplementary angles, therefore;
∠AOB + ∠COB = 180°
∠AOB = 180° - ∠COB
∴ ∠AOB = 180° - 76° = 104°
∠AOB = 104°
OA = OB = The radius of the circle
Therefore, ΔAOB = An isosceles triangle
∠OAB = ∠OBA by base angles of an isosceles triangle are equal
∠AOB + ∠OAB + ∠OBA = 180° by angle summation property
∴ ∠AOB + ∠OAB + ∠OBA = ∠AOB + ∠OAB + ∠OAB = ∠AOB + 2×∠OAB = 180°
∠OAB = (180° - ∠AOB)/2
∴ ∠OAB = (180° - 104°)/2 = 38°
∠TAC = ∠OAB = 38° by reflexive property
AOC is perpendicular to tangent PT at point C, by tangent to a circle property, therefore;
∠TCA = 90° and ΔTCA = A right triangle
∠TAC + ∠ATC + ∠TCA = 180° by angle sum property
∠ATC = 180° - (∠TAC + ∠TCA)
∴ ∠ATC = 180° - (38° + 90°) = 52°
Angle ATC = 52°
(b) In ΔABC, ∠ABC = Angle subtended by the diameter = 90°
∴ ΔABC = A right triangle
∠ABC and ∠TBC are supplementary angles, therefore;
∠ABC + ∠TBC = 180°
∠TBC = 180° - ∠ABC
∴ ∠TBC = 180° - 90° = 90°
∠TCB = 180° - (∠TBC + ∠ATC)
∴ ∠TCB = 180° - (90° + 52°) = 38°
The bearing of B from C = (360° - 38°) = 322°.