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How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=5, c=6?

1 Answer

Oct 22, 2016

(see below)

Explanation:

Based on the Pythagorean Theorem, we know that
#color(white)("XXX")a^2+b^2=c^2

$\to b = \sqrt{c^{2} - a^{2}} = \sqrt{6^{2} - 5^{2}} = \sqrt{11}$ $rarr b=sqrt(c^2-a^2)=sqrt(6^2-5^2)=sqrt(11)$
enter image source here

$sin = \frac{opposite}{hypotenuse} XX \to sin (A) = \frac{5}{6}$ $sin = "opposite"/"hypotenuse"color(white)("XX")rarr sin(A)=5/6$

$csc = \frac{hypotenuse}{opposite} XX \to csc (A) = \frac{6}{5}$ $csc = "hypotenuse"/"opposite" color(white)("XX")rarr csc(A)=6/5$

$cos = \frac{adjacent}{hypotenuse} XX \to cos (A) = \frac{\sqrt{11}}{6}$ $cos = "adjacent"/"hypotenuse"color(white)("XX")rarr cos(A)=sqrt(11)/6$

$sec = \frac{hypotenuse}{adjacent} XX \to sec (A) = \frac{6}{\sqrt{11}}$ $sec = "hypotenuse"/"adjacent" color(white)("XX")rarr sec(A)=6/sqrt(11)$

$tan = \frac{opposite}{adjacent} XX \to tan (A) = \frac{5}{\sqrt{11}}$ $tan = "opposite"/"adjacent"color(white)("XX")rarr tan(A)=5/sqrt(11)$

$cot = \frac{adjacent}{opposite} XX \to cot (A) = \frac{\sqrt{11}}{5}$ $cot = "adjacent"/"opposite" color(white)("XX")rarr cot(A)=sqrt(11)/5$

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