Law of Sines Goal: To solve triangles which aren’t necessarily

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Solution of Triangles SINE RULE. 22 angle dan 1 side are given e.g  A = 60 ,  B = 40  and side b = 8 cm then, side a & side c can be found using.

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Presentation transcript:

Law of Sines Goal: To solve triangles which aren’t necessarily right triangles ( find all sides & angles) C b a A B c or

: 0≤sinØ≤1 when 0˚≤Ø≤180˚ : There may be no answers, 1 answer, or 2 answers. (There are 2 angles--1 acute, 1 obtuse--that have the same sine.) : If there are 2 angles given, you can easily find the 3rd angle (the 3 angles of a triangle add up to 180˚).

: To change the decimal part of degrees to minutes, multiply the decimal part by 60. (._ _ means ‘hundredths’…we want ‘minutes’, which is 60ths.) : If the given angle Ø>90˚and there exists a solution, then there exists only 1 solution. : In order to use this formula, you must know at least 1 angle and its corresponding side OR 2 angles and 1 side.

: We found C by taking 180-(25˚26’+78˚). Ex: Determine the number of possible solutions. If a solution exists, solve the triangle. 1. a=13.7, A=25˚26’, B=78˚ : We found C by taking 180-(25˚26’+78˚). A= 25˚26’ a= 13.7 B= 78˚ b= C= c= 31.2 76˚34’ 31.0

: We need not go any farther. Ex: 2. a=33, A=132˚, b=50 A=132˚ a=33 Determine the number of possible solutions. If a solution exists, solve the triangle. 2. a=33, A=132˚, b=50 : We need not go any farther. A=132˚ a=33 B= b=50 C= c=

OR : We found C by taking 180-(sum of A+B). Ex: 1. a=125, A=25˚, b=150 Determine the number of possible solutions. If a solution exists, solve the triangle. 1. a=125, A=25˚, b=150 OR A= 25˚ a=125 B= b=150 C= c= : We found C by taking 180-(sum of A+B). 30˚28’ or 149˚32’ 124˚32’ or 5˚28’ 243.7 or 28.2’