1. START UP Day 42 A guy-wire connects the top of a pole to point on the ground at an angle of depression of 80º. On the ground, the wire is 4.5 ft from the base of the pole. (a) How long is the wire? (b) How tall is the pole? 80° 4.5 ft.

2. THE LAW OF SINES OBJECTIVE: Students will be able to demonstrate an understanding of the laws of sine and cosine. Essential Questions: How can we use the law of sines and the law of cosines to solve oblique triangles? How do we know which law should be used first? When might we expect two possible triangle solutions? Home Learning: p. 439 # 3, 6, 9, 13, 16, 18, 22, 23 28, 35 & 40 + Watch: http://youtu.be/qE3- 4ZfMCZQhttp://youtu.be/qE3- 4ZfMCZQ

3. THE LAW OF SINES HTTPS://YOUTU.BE/GAX_ILEQEJQ For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:

4. WHAT DO WE KNOW? Triangle problems break down into the following four cases. 1. Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines. http://youtu.be/It_bq5DzQ1Q

5. EXAMPLE 1 SOLVE for Δ ABC, given angle A = 70°, angle B = 80° and side a = 12 cm. * Please remember that angles are named with capital letters and the side opposite an angle is named with the same lower case letter.* Note: Use the Law of Sines whenever you have a complete PAIR!

6. Example 1 (con’t) AC B 70° 80° a = 12 c b The angles in a ∆ total 180°, so angle C = 30°. Set up the Law of Sines to find side b: A=70 degreesa= 12 B=80 degreesb= C=30 degreesc=

7. Example 1 (con’t) Set up the Law of Sines AGAIN to find side c: AC B 70° 80° a = 12 c b = 12.6 30° A=70 degreesa= 12 B=80 degreesb=12.576 C=30 degreesc=

8. THE TROUBLE WITH THE LAW OF SINES AND THE AMBIGUOUS CASE (SSA) When given SSA (two sides and an angle that is NOT the included angle), the situation may be ambiguous. The dimensions may not form a triangle, or there may be 1 or 2 triangles with the given dimensions. We first go through a series of tests to determine how many (if any) solutions exist. Remember that the inverse sine function has a limited range & will never “see” an obtuse angle!

9. THE AMBIGUOUS CASE (SSA) Situation I: The GIVEN ANGLE is OBTUSE NO Worries—It will be ONE or possibly- NONE! AB ? a b C = ? c = ? If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible. AB ? a b C = ? c = ? If a > b, then there is ONE triangle with these dimensions.

10. THE AMBIGUOUS CASE (SSA) Situation I: The Given Angle is obtuse - EXAMPLE Given a triangle with angle A = 120°, side a = 22 cm and side b = 15 cm, find the other dimensions. Since a > b, these dimensions are possible. To find the missing dimensions, use the Law of Sines: A B a = 22 15 = b C c 120°

11. THE AMBIGUOUS CASE (SSA) Situation I: Angle A is obtuse - EXAMPLE Angle C = 180° - 120° - 36.2° = 23.8° Use Law of Sines to find side c: A B a = 22 15 = b C c 120° 36.2° Solution: angle B = 36.2°, angle C = 23.8°, side c = 10.3 cm

12. THE AMBIGUOUS CASE (SSA) Situation II: The given angle is acute This is the tricky one—it all depends on the next angle that you can solve for—which depends on its given side length. If the next angle is greater than the given….expect 2 solutions, but it still could be a “no solution”. You just have to work it through. AB ? b C = ? c = ? a

13. THE AMBIGUOUS CASE (SSA) Situation II: Angle A is acute - EXAMPLE 1 Given a triangle with angle A = 40°, side a = 12 cm and side b = 15 cm, find the other dimensions. A B ? 15 = b C = ? c = ? a = 12 40° Is the next angle you can find bigger than the given? YES, because side b is longer than side a—so we will expect two solutions!

14. THE AMBIGUOUS CASE (SSA) Situation II: Angle A is acute - EXAMPLE 1 A B 15 = b C c a = 12 40° FIRST SOLUTION: Use the inverse sine function to determine Angle B as an acute angle – then use this angle to solve for the remaining values!

15. THE AMBIGUOUS CASE (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Use the Supplementary Angle to the original angle B (180- B). Angle B is obtuse –then solve for the remaining values! a = 12 A B 15 = b C c 40° 1st ‘a’ 1st ‘B’ In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary. Angle B = 180 - 53.5° = 126.5°

16. THE AMBIGUOUS CASE (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse a = 12 A B 15 = b C c 40° 126.5 ° Angle B = 126.5° Angle C = 180°- 40°- 126.5° = 13.5°

17. THE AMBIGUOUS CASE (SSA) Situation II: Angle A is acute - EXAMPLE 2 Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions. A B ? 10 = b C = ? c = ? a = 12 h 40° Since a > b, we know that Angle B will be less than angle A. Angle A will be acute. This triangle has only ONE possible solution.

18. THE AMBIGUOUS CASE (SSA) Situation II: Angle A is acute - EXAMPLE 2 Using the Law of Sines will give us the ONE possible solution: A B 10 = b C c a = 12 40°

19. THE AMBIGUOUS CASE - SUMMARY ONLY One Angle? ACUTE? If the next angle is bigger …you may have 2 solutions If the next angle is smaller… no worries! OBTUSE? No Worries! It will either be one solution or none If dec.<1 then you will have TWO! If dec.>1 then it can’t be done!

20. STAY THE COURSE! The course for a boat race starts at point A and proceeds in the direction S 52  W to point B, then in the direction S 40  E to point C, and finally back to A. Point C lies 8 kilometers directly south of point A. Approximate the total distance of the race course. 8km 40° 52°

21. A LITTLE GEOMETRY HELPS US THROUGH! Because lines BD and AC are parallel, it follows that angle C=40°. (Alternate Interior) The measure of angle B is 180  – 52  – 40  = 88 . Now that we have a complete pair, we can use the Law of Sines, Figure 6.10

22. SOLUTION Because b = 8, and The total length of the course is approximately Length  8 + 6.308 + 5.145 =19.453 kilometers.