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Unformatted text preview: 138.19◦ . At 3 3 3 this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for γ are ‘compatible’ with the given angle-side pair 11.2 The Law of Sines 765 (α, a) = (30◦ , 3) in that both choices for γ can fit in a triangle with α and both have a sine of 2 3 . The only other given piece of information is that c = 4 units. Since c > a, it must be true that γ , which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as 2 well. Thus have two triangles on our hands. In the case γ = arcsin 3 radians ≈ 41.81◦ , we find5 β ≈ 180◦ − 30◦ − 41.81◦ = 108.19◦◦. Using the Law of Sines with the angle-side opposite pair (α, a) and β , we find b ≈ 3 sin(108.◦19 ) ≈ 5.70 units. In the case γ = π − arcsin 2 radians 3 sin(30 ) ≈ 138.19◦ , we repeat the exact same steps and find β ≈ 11.81◦ and b ≈ 1.23 units.6 Both triangles are d...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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