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Section 7.2&7.3 - Section 7.2 The Law of Sines...

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Unformatted text preview: Section 7.2 The Law of Sines Oblique Triangles If none of the angles of a triangle is a right angle, the triangle is called oblique. To solve an oblique triangle is to find the length of its sides and all it angle measures. These in general are more difficult to solve and in certain cases have more than one answer. What do we need to know? We will always need to know the length of one side (since there are an infinite number of triangles who have the same angle measures). We will also need two other pieces of information, which can be any combination of angles and sides. ASA, SAA, SSA, SAS, SSS Law of Sines For any triangle with sides a,b,c and opposite angles ,, respectively, i.e. the angle is opposite a. sin()/a = sin()/b = sin()/c Why is this true? Partial Proof A C H B We know that sin() = h/a or that h = a*sin() We also have sin() = h/c or that h = c*sin() Thus we have a*sin() = c*sin() or sin()/c = sin()/a SAA and ASA If you know two angles you can always find the third by using ++=180 Once you have all 3 angles and one side you simply apply the law of sines. SSA This is the one that may have multiple one or no answers two. Do you see why? To solve these you use the law of sines. If there is no angle that makes it true then there is no solution. Otherwise you will get two angles that take it true, in which case you simply check the angles with ++=180 to see which works. Section 7.3 The Law of Cosines The Law of Cosines c2 = a2 + b2 2ab cos 2. b2 = a2 + c2 2ac cos 3. a2 = b2 + c2 2bc cos This will be used to solve SAS and SSS triangles. 1. Partial Proof a c H b If we place this triangle so that the side b is on the xaxis and the point at angle is at the origin we get, the point at is (a cos, a sin) and the point at is (0, b) Partial Proof a c H b Thus the distance formula gives us Thus the distance formula gives us c2 = (b a cos)2 + (0 a sin)2 = b2 2ab cos + a2cos2 + a2sin2 = b2 2ab cos + a2 ...
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