Area, Angle of Chords of a Circle

Date: 7/25/96 at 4:36:31
From: Wong Cheong Siong
Subject: Area, Angle of Chords of a Circle

Dear Dr. Math,

I have a circular measure question to ask you and I would appreciate 
it if you could solve them for me. The question is:

AB is a diameter of a circle whose centre is O. P is a point on the 
circumference such that the chord AP = 8cm and the chord BP = 6cm. 
Calculate (correct to 3 significant figures)

a) the values, in radians, of the angles PAB and POB,
b) the area of the sector bounded by OP, OB and the minor arc PB.

Thank You!

Date: 7/25/96 at 5:56:19
From: Doctor Anthony
Subject: Re: Area, Angle of Chords of a Circle

Since APB is the angle in a semicircle, it is a right angle, and we 
can use Pythagoras to find AB.  Thus AB^2 = 8^2 + 3^2
                            AB^2 = 64 + 9 = 73
                            AB = sqrt(73) = 8.544
                                Radius OB = 4.272      

Also tan(PAB) = 3/8 = 0.375

         PAB = 0.35877 radians
         POB = 2*PAB = 0.71754 radians  

We can say POB = 2*PAB because POB is the angle at the centre, and PAB 
the angle at the circumference, subtended by the chord PB.

The area of the sector POB is given by (1/2)*POB*radius^2

                                  = (1/2)*0.71754*4.272^2

                                  = 6.54756  cm^2

                           
-Doctor Anthony,  The Math Forum
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