6.1 Law of Sines +Be able to apply law of sines to find missing sides and angles +Be able to determine ambiguous cases

Oblique Triangles (A triangle with no right angles) A is acute A is obtuse h B c a h a c A C b C A b Now it’s time to derive the Law of Sines!

Law of Sines Reciprocals! If ABC is a triangle with sides a, b, and c, then Reciprocals!

Solving Oblique Triangles To solve oblique triangles you need one of the following cases 2 angles and any side (AAS or ASA) 2 sides and the angle opposite one of them (SSA) 3 sides (SSS) 2 sides and their included angle (SAS) We can apply the Law of Sines in the first two cases

Example 1 Use the Law of Sines to solve the triangle. (Find the missing information). C = 102.30 B = 28.70 b = 27.4 ft

Word Problem A pole tilts towards the sun at an 8° angle from the vertical, and it casts a 22 foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is 43°. How tall is the pole?

Ambiguous Cases (SSA) When you are given two sides and the angle opposite one of them, there are six possible cases that can occur. (also remember: h = bsinA) 1. 2. A is acute a < h no triangles A is acute a = h 1 triangle A b a h A a h b

More Cases 3. 4. A is acute A is acute a > b h < a < b 1 triangle A is acute h < a < b 2 triangles A b h a A b a h

More Cases 5. 6. A is obtuse A is obtuse a < b a > b no triangles A is obtuse a > b 1 triangle A a b A b a

Determine how many triangles there would be: A = 620, a = 10, b = 12 A = 980, a = 10, b = 3 A = 540, a = 7, b = 8

Example 3 Find the missing sides and angles of the triangle given the following information a = 22 in b = 12 in A = 420

Ex. 4: Finding two solutions Find two triangles for which a = 12, b = 31, and A = 20.50

Using Sine To Find Area 𝐴= 1 2 (𝑏𝑎𝑠𝑒)(ℎ𝑒𝑖𝑔ℎ𝑡) We know from our exploration before that the altitude, or height, is: ℎ=𝑏 sin 𝐴 ℎ=𝑎 sin 𝐵 ℎ=𝑎 sin 𝐶 Using the commutative property for multiplication, we get: Basically, area is half the product of the two sides times their included angle.

Find the Area of the Triangle Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102o.

HW pg 398-400 #1-7, 15-19 odd, 29, 32, 41, 42

You Try! Use the Law of Sines to solve the triangle. (Find the missing information). A = 100 a = 4.5 ft B = 600

You Try Find the missing sides and angles for the following information a = 15 b = 25 A = 850

You Try Use the Law of Sines to solve the triangle. (Find the missing information). A = 360 a = 8 in b = 5 in

Word Problem The course for a boat race starts at point A and proceeds in the direction S 52o W to point B, then in the direction S 40o E to point C, and finally back to point A. Point C lies 8 kilometers directly south of point A. Approximate the total distance of the race course.

+Be able to apply law of cosines to find missing sides and angles

Solving Oblique Triangles To solve oblique triangles you need one of the following cases 2 angles and any side (AAS or ASA) 2 sides and the angle opposite one of them (SSA) 3 sides (SSS) 2 sides and their included angle (SAS) We can apply the Law of Cosines in the last two cases

Law of Cosines

Alternate Form Solve each formula for the Cosine of the Angle

Given Three Sides (SSS) When you are given three sides of a triangle, it is best to find the angle that corresponds with the longest side first. Because the inverse cosine function has a range of [0, 180°], it will allow us to find the obtuse angle, if there is one. Inverse sine only has a range of [-90°, 90°], so we have to do extra work to find obtuse angles. Once we have one angle, we can use the Law of Sines to find a second, and then subtract from 180° to find the third.

Ex: Three Sides of a Triangle Find the angles for the triangle with side lengths a = 8 ft, b = 19 ft, and c = 14 ft

Two Sides and the Included Angle - SAS Find the remaining angles and the side of a triangle with the following information. A = 115° b = 15 cm c = 10 cm

Word Problem The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases in 60 feet. (the pitcher’s mound is not half way between home plate and second base) How far is the pitcher’s mound from first base?