Sunday, May 28, 2017

Chapter 27 - Additional Examples

Additional Example 1
In the figure (a) below, A, B, C and D are points on the circle. 
Compute the angles of the quadrilateral ABCD, and the angles between it's diagonals
Solution:
1. Consider ΔBPC. We get BPC = [180-(30+50)] =[180-80] = 100o. (∵ sum of interior angles of a triangle is 180o) This is marked in fig(b)
2. BPC and BPA form a linear pair. So BPA = 180 - BPC = 180 -100 = 80o.
3. BPC and APD are opposite angles, and are hence equal. So we get APD = BPC = 100o.
4 Similarly, BPA and DPC are opposite angles, and are hence equal. So we get DPC = BPA = 80o.
5. Consider arc BC in fig(c). It subtends BAC (= 35o) on the alternate arc. 
• The same arc subtends BDC on the alternate arc. So BDC = BAC = 35o.
6. Consider ΔPCD. We get PCD = [180-(80+35)] =[180-115] = 65o. (∵ sum of interior angles of a triangle is 180o)
7. Consider arc CD in fig(c). It subtends ∠CBD (= 30o) on the alternate arc. 
• The same arc subtends ∠CAD on the alternate arc. So CAD = ∠CBD = 30o.
8. Consider ΔPAD. We get ∠ADP = [180-(100+30)] =[180-130] = 50o. (∵ sum of interior angles of a triangle is 180o)
■ Thus we get all the angles of the quadrilateral ABCD. Since it is a cyclic quadrilateral, we can do a check:
(i) Sum of opposite angles BAD and BCD = 35 + 30 + 50 + 65 = 180o         
(ii) Sum of opposite angles ABC and ADC = 65 + 30 + 50 + 35 = 180o         

Additional Example 2
In the figure (a) below, AB is a diameter of the circle. 
CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Calculate AEB
Solution:
1.Draw OC and OD as shown in fig(b). Consider the ΔOCD
• OC = OD (∵ radii of the same circle)
• Given that chord CD is equal to the radius.
• So ΔOCD is an equilateral triangle. All it's interior angles are equal to 60o.
2. Draw AD as shown in fig(c)
3. Consider the arc CD. It has a central angle COD = 60o.
4. This same arc CD subtends CAD on the opposite arc. 
• So CAD = 12×60 = 30o.  
5. Now consider ADB. AB is a diameter and D is a point on the semi-circle. So ADB = 90o.
6. Consider ΔAED. The ADB that we considered above is an exterior angle of ΔAED
• Exterior angle of a triangle = sum of remote interior angles
• So ADB = EAD AED  90 = 30 AED  AED = 90 - 30 = 60o.
7. But AED and AEB are the same angles
• Thus the required AEB = 60o


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