CotA =5/12then sinA+cosa=?

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CotA =5/12then sinA+cosa=?

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Answer 1 · 2018-05-12T12:14:08

$sin A + cos A = \frac{17}{13}$ $sinA+cosA=17/13$

Explanation:

$cot A = \frac{5}{12}$ $cotA=5/12$
cotangent= adjecent/opposite
thus the hypotenuse is:
$h = \sqrt{25 + 144} = 13$ $h=sqrt(25+144)=13$
$sin A = \frac{12}{13}$ $sinA=12/13$
$cos A = \frac{5}{13}$ $cosA=5/13$
$sin A + cos A = \frac{12}{13} + \frac{5}{13} = \frac{17}{13}$ $sinA+cosA=12/13+5/13=17/13$

Answer 2 · 2018-05-13T01:33:39

$sin A + cos A = \pm \frac{17}{13}$ $sin A + cos A = +- 17/13$

Explanation:

Use trig identities to find sin A and cos A
${cos}^{2} A = \frac{1}{1 + {tan}^{2} A} = \frac{1}{1 + \frac{25}{144}} = \frac{144}{169}$ $cos^2 A = 1/(1 + tan^2 A) = 1/(1 + 25/144) = 144/169$
$cos A = \pm \frac{12}{13}$ $cos A = +- 12/13$
$cot A = \frac{5}{12}$ $cot A = 5/12$ --> A could be in Q. 1 or Q. 3.
${sin}^{2} A = 1 - {cos}^{2} A = 1 - \frac{144}{169} = \frac{25}{169}$ $sin^2 A = 1 - cos^2 A = 1 - 144/169 = 25/169$
$sin A = \pm \frac{5}{13}$ $sin A = +- 5/13$
If A is in Quadrant 1 -->
$sin A + cos A = \frac{12}{13} + \frac{5}{13} = \frac{17}{13}$ $sin A + cos A = 12/13 + 5/13 = 17/13$
If A is in Quadrant 3 -->
$sin A + cos A = - \frac{12}{13} - \frac{5}{13} = - \frac{17}{13}$ $sin A + cos A = - 12/13 - 5/13 = - 17/13$
Answer: $sin A + cos A = \pm \frac{17}{13}$ $sin A + cos A = +- 17/13$

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CotA =5/12then sinA+cosa=?

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