The obtuse angel B is such that tanB = -(5/12). How do you find the following values?

Question

sin B

cos B

sin 2B

cos 2B

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Answer 1 · 2017-09-28T02:14:00

Given that the angle is obtuse , $B in "quadrant II"$

Again $tanB=-5/12=>B~~157.3^@$

$sinB->+ve$

$cosB->-ve$

$sin2B->-ve" as " 2B in" quadrant IV"$

$cos2B->+ve" as " 2B in" quadrant IV"$

Now $cosB=1/secB=-1/sqrt(1+tan^2B)$

$=-1/(1+5^2/12^2)=-12/13$

$sinB=tanBxxcosB=(-5/12)xx(-12/13)=5/13$

$sin(2B)=(2tanB)/(1+tan^2B)==(-2xx5/12)/(1+(-5/12)^2)=-120/169$

$cos(2B)=(1-tan^2B)/(1+tan^2B)==(1-(-5/12)^2)/(1+(-5/12)^2)=119/169$