Question

3. In the diagram, AB is the tangent to the circle at P and PX is the diameter. Given that BPQ = 42°, find PÔX, PÂXQ and XPQ.

Answer 1

Answer:

yjkxxykykykuyyjkzyyjzgkx

Answer 2

TQS=90

0

(angle in a semicircle)

∠STQ=∠SQR (angle in the alternate segment)

In Δ TQR,

∠QRS=x=180−(∠STQ+∠TQR)

=180−[40+(90+40)]

=180−170=10

0

hope u hv understood........

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