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Trigonometry. Law of Sines. Area of Oblique Triangles. h. Area = ½ bh,. h = b sin A. So Area = ½ c b sin A. Area of Oblique Triangles. h. Area = ½ bh,. h = a sin B. So Area = ½ c a sin B. Area of Oblique Triangles. h. h = a sin C. Area = ½ bh,. So Area = ½ b a sin C.
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Trigonometry Law of Sines
Area of Oblique Triangles h Area = ½ bh, h = b sin A So Area = ½ c b sin A
Area of Oblique Triangles h Area = ½ bh, h = a sin B So Area = ½ c a sin B
Area of Oblique Triangles h h = a sin C Area = ½ bh, So Area = ½ b a sin C
Example AAS Do you remember the AAS theorem in Geometry? What does that theorem tell us?
Example ASA Did we study this in Geometry?
Example SSA Did we study this theorem in Geometry? What does that mean to us now? Let’s try an example and see.
Solve triangle ABC if a = 22 in, b = 12 in, A = 42˚. Example
Why? If this is such an easy problem to solve, why did we not have an SSA theorem in Geometry? Let’s try another example.
SSA with Obtuse Angle Is there something wrong with this triangle? Side opposite the largest angle MUST be the longest side. Therefore there is NO triangle.
Law of Sines – Acute Angle Try this! What if a = 15, b = 25, and A = 85˚? 15 25 h 85˚ h = 25 sin 85˚ = 24.9 If side opposite is less than height; there is NO triangle.
Law of Sines – Acute Angle Try this! Find the triangle for which a = 12, b = 31, and A = 20.5˚. Notice: h = 31 sin 20.5˚ = 10.86 12 31 If side opposite is greater than height; there are 2 triangles. 20.5˚ There are 2 triangles!
Law of Sines C C’ A B A B’
Summary Ambiguous CaseSSA In any triangle when you are given a, b and A: • A obtuse • Side a < Side b – NO triangle • Side a > Side b – ONE triangle • A acute • Side a > Side b – ONE triangle • Side a < Side b • Find height (h = b sin A) • Side a < h – NO triangle • Side a = h – ONE triangle (right triangle) • Side a > h – TWO triangles