I suppose you know how to get to :

Angle CPQ (or APQ) = sin-¹(3.5/5)

And that you've also applied the converse of the property ' the perpendicular bisector of a chord passes through the centre'

Now realise that triangle CPQ and APQ are right angled triangles (because angle CQP and AQP are 90° as deduced from the property above)

Since the angle sum of a triangle is 180°,
then angle CPQ + angle QCP
= 180° - angle CQP
= 180° - 90°
= 90°

(Likewise for angle APQ)


So, angle CPQ and angle APQ must be < 90° for the above equality to hold true

(we don't have negative angles in triangles and the two non-right angles in a right angled triangle are always acute)

Therefore, even though sin-¹(3.5/5) could be ≈ 0.775396 rad (44.42700°) or ≈ 2.366195 rad (135.57299°), we reject the obtuse angle since we have already deduced that angle CPQ and angle APQ are acute.

It follows that angle CPA
= 2 x angle CPQ or 2 x angle APQ
≈ 2 x 0.775396 rad
= 1.550792 rad
= 1.55 rad (3 s.f)

thanks . it didn’t occur to me to draw the perpen bisector .

Welcome. Oh by the way, replace all the Q's in the above comment with the letter you would use to name the midpoint of AC since Q is on the arc and is not the midpoint of the segment AC.

thanks...

Hi J , if qn is : 3sinx + 4cosx = 2 ,
besides R formula , is there another method ?

Yes, but it's not taught at your level and is out of the syllabus.

https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

thanks . are u a tutor

Yup

Another alternative method :


3sinx + 4cosx = 2
So 3sinx = 2 - 4cosx

Square both sides,

9sin²x = 4 - 16cosx + 16cos²x

9 - 9cos²x = 4 - 16cosx + 16cos²x

25cos²x - 16cosx - 5 = 0


Using quadratic formula,

cosx = 8/25 ± 3√21 /25

x = cos-¹(8/25 + 3√21 /25)
or x = cos-¹(8/25 - 3√21 /25)

x ≈ 29.55(rej), 330.45°, 389.55°(rej), 690.45°,...

or

x ≈ 103.31,° 256.69°(rej), 463.31°, 616.69°(rej),...


Note that we squared the expression at first. So not all values of x found will satisfy the initial equation

3sinx + 4cosx is positive (>0)


When cosx is negative, x is in the 2nd or 3rd quadrant. But when the x is in the 3rd quadrant, both cosx and sinx are negative.

So this would result in a negative value of 3sinx + 4cosx . So all 3rd quadrant values of x are rejected.

For the rest, you'll have to check and see whether the 1st, 2nd and 4th quadrant values satisfy the equation.