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

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

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 , $B \in quadrant II$ $B in "quadrant II"$

Again $tan B = - \frac{5}{12} \Rightarrow B \approx {157.3}^{\circ}$ $tanB=-5/12=>B~~157.3^@$

$sin B \to + v e$ $sinB->+ve$

$cos B \to - v e$ $cosB->-ve$

$sin 2 B \to - v e as 2 B \in quadrant IV$ $sin2B->-ve" as " 2B in" quadrant IV"$

$cos 2 B \to + v e as 2 B \in quadrant IV$ $cos2B->+ve" as " 2B in" quadrant IV"$

Now $cos B = \frac{1}{sec B} = - \frac{1}{\sqrt{1 + {tan}^{2} B}}$ $cosB=1/secB=-1/sqrt(1+tan^2B)$

$= - \frac{1}{1 + \frac{5^{2}}{12^{2}}} = - \frac{12}{13}$ $=-1/(1+5^2/12^2)=-12/13$

$sin B = tan B \times cos B = (- \frac{5}{12}) \times (- \frac{12}{13}) = \frac{5}{13}$ $sinB=tanBxxcosB=(-5/12)xx(-12/13)=5/13$

$sin (2 B) = \frac{2 tan B}{1 + {tan}^{2} B} = = \frac{- 2 \times \frac{5}{12}}{1 + {(- \frac{5}{12})}^{2}} = - \frac{120}{169}$ $sin(2B)=(2tanB)/(1+tan^2B)==(-2xx5/12)/(1+(-5/12)^2)=-120/169$

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

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

sin B

cos B

sin 2B

cos 2B

1 Answer

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

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