✔ 最佳答案
10) With △PQR and △PRT being a pair of congurent isos. triangles (SAS):
Let ∠QPR = TPR = x, then
∠PQR = ∠PRQ = ∠PRT = ∠PTR = (180 - x)/2
Hence ∠QRT = ∠PRQ + ∠PRT = 180 - x
Thus ∠QRS = x = ∠QPR
Hence ST is a tangent to the circle at R (Converse of angle in the same segment)
13a) Join AC, then ∠BAC = 72 (∠ in alt. segment)
Let ∠TAB = x, then ∠BAC = x (∠ in alt. segment)
In △TAC, x + 72 + x + 38 = 180
2x = 70
x = 35
b) ∠ABC = 180 - ∠ACB - ∠BAC = 180 - 35 - 72 = 73
7) ∠DOB = 180 - 42 = 138 (Opp. ∠s, cyclic quad.)
∠BAD = 1/2 x ∠DOB = 69 (∠ at centre = Twice ∠ at circum)
8a) EF touches both circles at T
Join PT and QT, both of them are perp. to EF.
Thus P, T, Q is a straight line.
b) Radius of smaller circle = 6 cm
Hence radius of larger circle = 10 cm
Thus PQ = 16 cm