Question

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Answer 1

a. The required proof is obtained from triangles OAB and ABC formed by joining AB and that we have;

A
`\hat O`B + 2∠C = 180°, 2·A
`\hat P`B + 2∠C = 180°

∴ A
`\hat O`B = 2·A
`\hat P`B

b. i) P
`\hat S`S = P
`\hat R`Q, given that angles subtended by the same arc or chord are equal, therefore, in ΔPQS, we have;

`\left | PS \right |` =
`\left |PQ \right |`

ii) S
`\hat Q`P = 45°,
`S\hat R Z` =90°

Explanation:

a. The given parameters are;

The center of the circle is point O

Points on the circumference of the circle = A, B, and P

Required to be proved, A
`\hat O`B = 2·A
`\hat P`B

Let ∠O represent A
`\hat O`B and let ∠P' represent A
`\hat P`B

We draw a line from the center O to the point P, and a line joining points A and B on the circumference of the circle

In ΔOAB, we have;

∠O + 2∠C = 180° (The sum of the interior angles of a triangle)

In ΔAPB, we have;

∠P' + ∠(C - a) + ∠(P' + C + a) = 180°

∴ 2·∠P' + 2·∠C = 180°

Therefore, by addition property of equality, we get;

∠O = 2·∠P'

Therefore;

A
`\hat O`B = 2·A
`\hat P`B

b. i) The given parameters are;

Points on the circle = P, Q, R, and S

P
`\hat Q`S = P
`\hat R`Q

According to circle theory, the angles which an arc or chord subtends in a given segment are equal, therefore;

P
`\hat S`S = P
`\hat R`Q

Therefore, P
`\hat S`S = P
`\hat Q`S by transitive property of equality

P
`\hat S`S and P
`\hat Q`S are base angles of ΔPQS, given that P
`\hat S`S = P
`\hat Q`S, we have;

ΔPQS is an isosceles triangle with base QS and therefore, the sides PS and PQ are the equal sides

Therefore, we have;

`\left | PS \right |` =
`\left |PQ \right |`

ii) Given that SQ is the diameter of the circle, we have by circle theorem, the angle subtended on the circumference by the diameter = 90°

∴
`S\hat PQ` = 90°

From (i), we have that P
`\hat S`S = P
`\hat Q`S, therefore, in triangle ΔPQS, we have;

`S\hat PQ` + P
`\hat S`S + S
`\hat Q`P = 180°

Therefore;

90° + P
`\hat S`S + S
`\hat Q`P = 180°

P
`\hat S`S + S
`\hat Q`P = 180° - 90° = 90°

P
`\hat S`S = P
`\hat Q`S, therefore, P
`\hat S`S + S
`\hat Q`P = 2·S
`\hat Q`P

P
`\hat S`S + S
`\hat Q`P = 2·S
`\hat Q`P = 90°

S
`\hat Q`P = 90°/2 = 45°

S
`\hat Q`P = 45°

Similarly, given that SQ is the diameter, of the circle the angle
`S\hat R Q` formed by jointing S to Q is 90°

`S\hat R Q` = 90°

`S\hat R Q \ and \ S\hat R Z` are angles on a straight line and are therefore, supplementary, therefore;

`S\hat R Z` = 180° -
`S\hat R Q`

`S\hat R Z` = 180° - 90° = 90°

`S\hat R Z` =90°.