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did i get this right
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get what right theres nothing :P
they didnt show a triangle
ok so how do we do that
so is this the equation i have so far: 4^2+b^2=5^2
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@sailor wrote:
Then do the same with 5^2
so i got 16+b^2=25
b^2=9
3
so the answer is 3
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the answer isnt 3
i put 90 before and i got that wrong
mmh
\[\sin Q=\frac{ 4 }{ 5},\]\[\cos Q=\sqrt{1-\sin ^2Q}\]\[=\sqrt{1-(\frac{ 4 }{ 5 })^2}\]\[=\sqrt{1-\frac{ 16 }{ 25 }}\] \[=\sqrt{\frac{ 25-16 }{ 25 }}\]\[=\sqrt{\frac{ 9 }{ 25}}\]\[=\frac{ 3 }{ 5 }\] \[Q \le 90,\cos Q \ge 0\] as P+Q=90 P=90-Q \[\cos P=\cos (90-Q)\]\[=\sin Q=\frac{ 4 }{ 5}\] \[\cos P+\cos Q=\frac{ 4}{ 5 }+\frac{ 3}{ 5}=\frac{ 4+3 }{ 5 }=?\]
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