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In Triangle ABC, Cos B = (9/12) .....find Sin B. Answer 15/9 12/9 7.94/12 9/12
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note that cos^2 B + sin^2 B = 1
7.94/12
\[opposite=\sqrt{12^2-9^2}\]\[\sqrt{144-81}\] =7.937 sinB =7.937/12
Don't think that we need a diagram buddy
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u cud instead use cosB^2 + sinB^2 =1. . .sinB= sqrt(1-(9/12)^2)))
\[\sin x=\sqrt{1-(9/12)^2} \sin x=\sqrt{144-81/144} sinx =\sqrt{63/144} sinx=\sqrt{0.43}\]
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