Question #57a6c

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Question #57a6c

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Answer 1 · 2017-01-26T22:46:07

see below

Explanation:

Since $tan α = \frac{o}{a} = \frac{20}{21}$ $tan alpha =o/a =20/21$ using pythagorean theorem we have

$c^{2} = 20^{2} + 21^{2}$ $c^2=20^2+21^2$

$c = \sqrt{20^{2} + 21^{2}} = \sqrt{841} = 29$ $c=sqrt(20^2+21^2)=sqrt841=29$ Therefore, $cos α = \frac{21}{29}$ $cos alpha = 21/29$ and $sin α = \frac{20}{29}$ $sin alpha = 20/29$

We also need to half $180^{\circ} < α < 270^{\circ}$ $180^@< alpha < 270^@$ hence,

$\frac{180}{2^{\circ}} < \frac{α}{2} < \frac{270^{\circ}}{2}$ $180/2^@< alpha/2 < 270^@/2$

$90^{\circ} < \frac{α}{2} < 135^{\circ}$ $90^@ < alpha/2 < 135^@$ --> Quadrant II

$sin (\frac{α}{2}) = \sqrt{\frac{1}{2} (1 - cos α)} = \sqrt{\frac{1}{2} (1 - \frac{21}{29})}$ $sin (alpha/2) = sqrt(1/2(1-cosalpha))=sqrt(1/2(1-21/29)$ = $sqrt(1/2((29-21)/29)) =sqrt(1/2(8/29))=sqrt(4/29) = 2/sqrt29 :. = (2sqrt29)/29$

$cos (alpha /2)= -sqrt(1/2(1+cosalpha))=-sqrt(1/2(1+21/29))= - sqrt(1/2((29+21)/29) ) = -sqrt(1/2(50/29) )= -sqrt(25/29) = -5/sqrt29 :. = (-5sqrt29)/29$

$tan (alpha / 2) = sin(alpha/2)/cos(alpha/2) = (2/sqrt29)/(-5/sqrt29) = 2/sqrt29 * sqrt29 / -5$

$= 2/cancel sqrt29 * cancel sqrt29 / -5 :. = -2/5$

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Question #57a6c

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