McDougal Littell Jurgensen Geometry: Student Edition Geometry
McDougal Littell Jurgensen Geometry: Student Edition Geometry
5th Edition
ISBN: 9780395977279
Author: Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Publisher: Houghton Mifflin Company College Division
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Chapter 9.2, Problem 5CE
To determine

To find:The congruent angles in given diagram as well as the pairs of complementary angles. Also to find the pair of supplementary angles.

Expert Solution & Answer
Check Mark

Answer to Problem 5CE

(i) Congruent angles pair:

(i) APO,BPO (ii) POB,POA (iii) PBO,PAO

(ii) Complementary angles pair: (i) POAand APO (ii) POB and BPO .

(iii) Supplementary angles pair: PBO,PAO

Explanation of Solution

Giveninformation: The below diagram is given,

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 9.2, Problem 5CE

Concept used:Following properties are used here,

  1. In a circle, any line from center of circle to the point of contact of tangent is always perpendicular to tangent.
  2. The line passing from the center and intersecting point of two tangents to its circle, always bisect the two angles, forming on the center and intersecting point.
  3. In a right triangle, remaining two angles, except right angle, are complementary angles

Also, if the sum of two angles is 90° , so these are complementary and if their sum is 180° , so these angles are supplementary.

Calculation:So, in given figure as OAAP , so PAO=900 and thus ΔAPO is a right triangle. Similarly as OBBP , so PBO=900 . Hence, ΔBPO is a right triangle. Thus, using first property, PBO=PAO=900 . So, these two angles are congruent.

Also, the sum of these two angles will be 180 degree, so these two angles are supplementary also.

Further, in right triangle ΔAPO , as angle A is 90 degree, so by angle sum property of triangle,

  POA+APO=900 , hence these two angles are complementary to each other. Similarly in right triangle ΔBPO , POB+BPO=900 . So, these two are also complementary angles. Also, using third property, another sets of congruent angles are POB=POA and APO=BPO .

Conclusion: Thus, it is concluded that pairs of congruent angles are

(i) APO,BPO (ii) POB,POA (iii) PBO,PAO .

Pairs of complementary angles are

(i) POAand APO ii) POB and BPO .

Further only pair of supplementary angles is PBO,PAO .

Chapter 9 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 9.1 - Prob. 11CECh. 9.1 - Prob. 1WECh. 9.1 - Prob. 2WECh. 9.1 - Prob. 3WECh. 9.1 - Prob. 4WECh. 9.1 - Prob. 5WECh. 9.1 - Prob. 6WECh. 9.1 - Prob. 7WECh. 9.1 - Prob. 8WECh. 9.1 - Prob. 9WECh. 9.1 - Prob. 10WECh. 9.1 - Prob. 11WECh. 9.1 - Prob. 12WECh. 9.1 - Prob. 13WECh. 9.1 - Prob. 14WECh. 9.1 - Prob. 15WECh. 9.1 - Prob. 16WECh. 9.1 - Prob. 17WECh. 9.1 - Prob. 18WECh. 9.1 - Prob. 19WECh. 9.1 - Prob. 20WECh. 9.1 - Prob. 1ECh. 9.1 - Prob. 2ECh. 9.1 - Prob. 3ECh. 9.1 - Prob. 4ECh. 9.1 - Prob. 5ECh. 9.2 - Prob. 1CECh. 9.2 - Prob. 2CECh. 9.2 - Prob. 3CECh. 9.2 - Prob. 4CECh. 9.2 - Prob. 5CECh. 9.2 - Prob. 1WECh. 9.2 - Prob. 2WECh. 9.2 - Prob. 3WECh. 9.2 - Prob. 4WECh. 9.2 - Prob. 5WECh. 9.2 - Prob. 6WECh. 9.2 - Prob. 7WECh. 9.2 - Prob. 8WECh. 9.2 - Prob. 9WECh. 9.2 - Prob. 10WECh. 9.2 - Prob. 11WECh. 9.2 - Prob. 12WECh. 9.2 - Prob. 13WECh. 9.2 - Prob. 14WECh. 9.2 - Prob. 15WECh. 9.2 - Prob. 16WECh. 9.2 - Prob. 17WECh. 9.2 - Prob. 18WECh. 9.2 - Prob. 19WECh. 9.2 - Prob. 20WECh. 9.2 - Prob. 21WECh. 9.2 - 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