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Mathematics 13 Online
OpenStudy (anonymous):

PLEAAASSEEE help Pre Calc: thank you so much picture posted below xoxo vvv

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

you first have to use the pythagorean theorem to find the length of AB

jimthompson5910 (jim_thompson5910):

AC^2 + BC^2 = AB^2 36^2 + 24^2 = x^2 1296 + 576 = x^2 keep going to solve for x

OpenStudy (anonymous):

x=+/- 12 to the power of 13

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

you mean \[\Large x = \pm 12\sqrt{13}\] right?

OpenStudy (anonymous):

yes :) @jim_thompson5910

jimthompson5910 (jim_thompson5910):

since AB is the length and x is the length, we only consider positive lengths so \[\Large x = 12\sqrt{13}\] which means \[\Large AB = 12\sqrt{13}\]

jimthompson5910 (jim_thompson5910):

now use the definition of sine, plug in the lengths, and simplify sin(angle) = opposite/hypotenuse sin(A) = 24/(12*sqrt(13)) sin(A) = 2/(sqrt(13)) sin(A) = (2*sqrt(13))/(sqrt(13)*sqrt(13)) sin(A) = (2*sqrt(13))/(13)

OpenStudy (anonymous):

ughhh thank you so much! FINALLY I understand this nonsense.

jimthompson5910 (jim_thompson5910):

that's great

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