Area, Angle of Chords of a CircleDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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