If p is also on the graph of r 2 2 sin then p has a

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Unformatted text preview: sin(45◦ ) = 7 3 6 ≈ 5.72 units. Now that b 7 sin(120 we have two angle-side pairs, it is time to find the third. To find γ , we use the fact that the sum of the measures of the angles in a triangle is 180◦ . Hence, γ = 180◦ − 120◦ − 45◦ = 15◦ . To find c, we have no choice but to used the derived value γ = 15◦ , yet we can minimize the propagation of error here by using the given angle-side opposite pair (α, a). The Law of Sines ◦ ◦ ◦) gives us sin(15 ) = sin(120 ) so that c = 7 sin(15◦ ) ≈ 2.09 units. We sketch this triangle below. c 7 sin(120 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β , we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦ . As in the previous example, we are forced to use a derived value in our ◦computations since the only ◦ angle-side pair available is (γ, c). The Law of Sines gives sin(85 ) = sin(65 ) . After the usual a 5.25 ◦ rearrangement, we get a = 5.25 sin(85 ) ≈ 5.77 units. To find b we use the angle-side pair (γ,...
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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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