maths 題題題目2條

Draw a line joining AB, and also joining OA and OB, where O is the centre of the circle.

Now, as AR and BR are the tangents to the circle, so AR = BR (tangent property)

Therefore, ∠RAB = ∠RBA = (180* - 64*)/2 = 58*

Consider the quadrilateral OARB,

As OA = OB = radius of the circle,

so, ∠OAB = ∠OBA

Let x be ∠AOB

So, ∠OAB = ∠OBA = (180* - x)/2 = 90* - x/2

But note that AR is a tangent and OA is a radius,

so, ∠OAR = 90* (line joining centre to tangent perp. tangent)

So, 58* + (90* - x/2) = 90*

x = 116*

∠AOB = 2∠ACB (∠ at centre twice ∠ at circumference)

Therefore, ∠ACB = 116*/2 = 58*


4.a. AQ = PQ - PA = (11 - x) cm

PB = PA = x cm (tangent property)

BR = RP - PB = (9 - x) cm


b. CR = BR = (9 - x)cm (tangent property)

QC = AQ = (11 - x) cm (tangent property)

As QR = QC + CR

So, 10 = (11 - x) + (9 - x)

2x = 10

x = 5