Chapter 7 – UQ: How do you solve for missing sides and angles in a non-right triangle?

 EQ: How is the Law of Sines used to solve non- right triangles?

Any triangle can be divided into two right triangles.

Side a is always Opposite angle A Side b is always Opposite angle B Side c is always Opposite angle C A B C bc a

 Find the sin B  Find the sin C A B C hbc

Since sin B And the sin C By the transitive property A B C hbc Repeat the procedure this time drawing the line from B You can use any two parts at a time

A BC Given: <A = 23° 15’ <B = 18° a = 28 Find: <C b = c = Since you use two parts at a time, you will need a set of opposites a and <A, b and <B, or c and <C Example 1A

 So, how would we get started with a problem like this?

A BC 6b 8 30 ° Since triangles have 180 ° And: (cos, sin) (-, +) (+, +) (-, -) (+, -) Sin is positive in I and II: Is it possible that <A might have a different value?

A BC 6b 8 30 ° But: Since triangles have 180 ° And: (cos, sin) (-, +) (+, +) (-, -) (+, -) Is it possible that <A might have a different value? A BC 6b 8 30 ° If <A = 41.8, then < B =108.2 And b = 11.4 since sine is positive in quadrant 1 and 2 then <A could = =138.2 <B = =

A BC ° A BC ° Two tests: 1)If the angles in the triangle add up to 180 2)The smallest side is opposite the smallest angle and the largest is opposite the largest.

 Example 3: The triangle we solved in Ex 4A had two different solutions. The triangle pictured below gives the same information SSA, is it possible for this one to have two solutions?

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 EQ: When and how is the Law of Cosines used to solve non-right triangles?

What do we need to know about a triangle to be able to use Law of Sines? Circle the information combinations able to be solved by Law of Sines. AAS SAS ASA SSA AAA SSS

FIRST remember given the unit circle: (cos x, sin x) 1 (b cos x, b sin x) b sin 2 A + cos 2 A = 1 SECOND if the circle has a radius of b units: Cos x Sin x x x

Given triangle ABC A(b cos C, b sin C) C(0,0) B(a,0) a b c Use the distance formula: To find the length of side c c 2 =(bcos C – a) 2 + (b sin C – 0) 2 c 2 = b 2 cos 2 C – 2ab cos C + a 2 +b 2 sin 2 C c 2 = a 2 + b 2 sin 2 C + b 2 cos 2 C -2ab cosC c 2 = a 2 + b 2 (sin 2 C + cos 2 C) – 2 ab cos C c 2 = a 2 + b 2 – 2ab cos C With points as indicated You can memorize this in three formats or one general one: Side 1 2 = Side Side 3 2 – 2 Side 2 Side 3 cos Angle 1

Example 1A Given b = 100, c = 75, m <A = 170 B A C

Example 3A Given a = 12, b = 13, c = 14 A B C

Example Given m <B = 120, m <C = 30, a = 16 Do you have a choice in your approach to this problem? A B C

 Which techniques should you use to solve word problems involving non-right triangles?

 How do you know when to use Law of Sines and when to use Law of Cosines? Do you sometimes, always, or never have a choice?

A pole leans away from the sun at an angle of 7º from the vertical. When the angle of elevation of the sun is 51º, the pole casts a shadow 47 feet long on level ground. How long is the pole? x 7° 51° 47

A hill is inclined 15º to the horizontal. A 40-ft. pole stands at the top of the hill. How long a rope will it take to reach from the top of the pole to a point 68 feet downhill from the base of the pole? 15° 40’ 68’ x

 Points A and B are on opposite sides of a lunar crater. Point C is 62m from A. The measure of <BAC is determined to be 102 degrees and the measure of <ACB is determined to be 52 degrees. What is the width of the crater? A B 62’ C 102° 52° x

A vertical antenna is mounted on top of a 60 ft. building. From a point on the level ground 35 ft. from the base of the pole, the antenna subtends an angle of 3.5º. Find the length of the antenna 60’ 3.5° 35’ x

One airplane leaves in a direction of 25° from the airport and flies for 2 hours at a speed of 200 mph. A second plane leave the same airport one hour later in a direction of 200° from the airport and flies for 1 hour at a speed of 300 miles per hour. How far apart are the planes when the second plane has been flying for one hour? N S EW 400 mi 300 mi x 25° 200° 175° If it says on a bearing, the measurement is from North—it does not change the angle between the flight paths or ships paths