Question

Help with a trigonometric question, please! It's posted in the comments.

Answer 1

@misty1212

Answer 2

\(\bf 0<\theta<90^o\textit{means the 1st quadrant, means sine and cosine are both positive}
\\ \quad \\
sin(\theta)=\cfrac{21}{29}\qquad sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad thus
\\ \quad \\
sin(\theta)=\cfrac{21}{29}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b}{c}\to \cfrac{b=21}{c=29}
\\ \quad \\
\textit{using the pythagorean theorem }\to c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\\ \quad \\
\pm\sqrt{29^2-21^2}=a\)

what would that give youu for "a" then? recall "a" = "adjacent side"

Answer 3

a=20

Answer 4

yeap so now we know that the adjacent side is 20
notice that, the root could give a -20 or a +20, so it couild really be either
because either one squared, will give 400 anyway
BUT, we know the angle is in the 1st quadrant, and on the 1st quadrant, cosine is positive
and also is the adjacent side, thus we'll use the +20 then
thus
\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\to \cfrac{a}{c}\to \cfrac{20}{29}\)

Answer 5

So the answer is C. Thanks! It was really easy to follow you and understand what you were trying to explain.

Answer 6

:)