Warm up   A 5.2 m ladder leans against a wall. The bottom of the ladder is 1.9 m from the wall. What angle does the ladder make with the ground (to the nearest degree)? Cos-1(1.9/5.2) = 69o

Objective: To use the Law of Sines in order to solve oblique triangles

Consider the first category, an acute triangle (, ,  are acute).

Create an altitude, h.

Theorem Law of Sines

The Law of Sines is used when we know any two angles and one side or when we know two sides and an angle opposite one of those sides.

Applying the Law of Sines The Law of Sines may be used when the known parts of the triangle are: 1. one side and two angles (SAA), (ASA) 2. two sides and an angle opposite one of the sides (SSA)

Example: In triangle ABC, angle A = 106 o, angle B = 31o and side a = 10 cm. Solve the triangle ABC by finding angle C and sides b and c.(round answers to 1 decimal place). C= 43 degrees c = 7.1 cm

Solution Use the fact that the sum of all three angles of a triangle is equal to 180 o to write an equation in C. A + B + C = 180 o Solve for C. C = 180 o - (A + B) = 43 o

Solution (cont’d) Use sine law to write an equation in b. a / sin(A) = b / sin(B) Solve for b. b = a sin (B) / sin(A) = (approximately) 5.4 cm Use the sine law to write an equation in c. a / sin(A) = c / sin(C) Solve for c. c = a sin (C) / sin(A) = (approximately) 7.1 cm

Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to the triangles below, that each triangle has a height of h = b sin A. A is acute. A is obtuse.

Area of a Triangle - SAS SAS – you know two sides: b, c and the angle between: A Remember area of a triangle is ½ base ● height Base = b Height = c ● sin A  Area K= ½ bc(sinA) A B C c a b h Looking at this from all three sides: K = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

Example: Find the area of given a = 32 m, b = 9 m, and

Area of a Triangle You can also find the area if you know one side and 2 angles based on the Law of Sines. so, substitute for b in the last equation, K = ½ bc(sinA) gives you

Area of a Triangle Find the area of triangle JKL if j=45.7, K=111.1o, and L=27.3o. 673.0 units sq.

Law of Sines practice http://www.emathematics.net/trigonometria. php?tr=5

Warm up Using the triangle above, A = 50o, B = 65o and a = 12.  Solve the triangle. / C = 65o b = 14.2 c = 14.2  

Lesson 5-7 Law of Sines the Ambiguous Case Objective: To determine whether a triangle has zero, one or two solutions and solve using the Law of Sines.

The Ambiguous Case – SSA In this case, you may have information that results in one triangle, two triangles, or no triangles.

SSA – No Solution Two sides and an angle opposite one of the sides.

By the law of sines,

Therefore, there is no value for  that exists! No Solution! Thus, Therefore, there is no value for  that exists! No Solution!

SSA – Two Solutions

By the law of sines,

So that,

Case 1 Case 2 Both triangles are valid! Therefore, we have two solutions.

Case 1

Case 2

For SSA Triangles: If A < 90° If A ≥ 90° a < b a ≥ b 1 Solution 1. a < b(sin A) No Solution a = b(sin A) 1 Solution a > b(sin A) 2 Solution a ≥ b 1 Solution If A ≥ 90° a ≤ b No Solution a > b 1 Solution

Practice Solve the triangle: A = 42°, a = 11, and b = 6