In the diagram, the points A, B, C and D lie on the circumference of a circle, centre O. The line ST is the tangent at A and the radius AO, when produced, meets DC at P. Given that angle ADC = 75° and angle APD = 50°.

Here's the solution to question 11: a) Step 1: Calculate ABC. ABCD is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral sum to 180^. ABC + ADC = 180^ Given ADC = 75^. ABC = 180^ - 75^ ABC = 105^ Step 2: Calculate SAD. In APD: The sum of angles in a triangle is 180^. PAD + APD + ADP = 180^ Given APD = 50^. Since D, P, C are collinear, ADP = ADC = 75^. PAD + 50^ + 75^ = 180^ PAD = 180^ - 125^ PAD = 55^ The radius AO is perpendicular to the tangent ST at point A. SAO = 90^ SAD = SAO - DAO Since P lies on the extension of AO, DAO = PAD = 55^. SAD = 90^ - 55^ SAD = 35^ Step 3: Calculate POC. In OAD, OA and OD are radii, so OAD is an isosceles triangle. ODA = OAD = 55^ Now, find ODC: ODC = ADC - ODA ODC = 75^ - 55^ ODC = 20^ In ODC, OC and OD are radii, so ODC is an isosceles triangle. OCD = ODC = 20^ In OPC: The sum of angles in a triangle is 180^. POC + OPC + OCP = 180^ OPC is the same as APD = 50^. OCP is the same as OCD = 20^. POC + 50^ + 20^ = 180^ POC = 180^ - 70^ POC = 110^ b) Step 1: Express EHK in terms of x. The diagonals of a rhombus intersect at right angles. Therefore, EKH = 90^. In EHK, the sum of angles is 180^. HEK + EHK + EKH = 180^ Given HEK = x^. x^ + EHK + 90^ = 180^ EHK = 180^ - 90^ - x^ EHK = (90 - x)^ Step 2: Express FGH in terms of x. The diagonals of a rhombus bisect the angles of the rhombus. Given HEK = x^. This means HEK is half of FEH. FEH = 2 × HEK FEH = 2x^ Opposite angles of a rhombus are equal. FGH = FEH FGH = 2x^ Send me the next one 📸