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Mathematics 24 Online
OpenStudy (anonymous):

1. Draw a 30-60-90 triangle. Label all angle measures and side relationships. Using the side relationships from the figure, show that the following trigonometric identities hold true for the given angles:

OpenStudy (anonymous):

tan 60=sin 60/cos 60

OpenStudy (anonymous):

i know it is true, but why

OpenStudy (anonymous):

\[\sin ^{2}(30)+\cos ^{2}(30)\]

OpenStudy (anonymous):

ummmm im not doin tricanometry or geometry so im sorry :(

OpenStudy (anonymous):

and this too i got the triangle and the sides

OpenStudy (anonymous):

its alright :)

OpenStudy (ikram002p):

now from the graph you would know cos 60 =A/C sin 60=B/C tan 60 =B/C |dw:1407110185239:dw| now sin 60 / cos 60 = B/C *C/A=B/A=tan 60 does this make sense to u ?

OpenStudy (anonymous):

nop

OpenStudy (ikram002p):

i made a typo on the graph :P it should be like this |dw:1407110267063:dw| ok ill do from skratch

OpenStudy (ikram002p):

ok u have this triangle with sides A,B,C with me so far? |dw:1407110355925:dw|

OpenStudy (anonymous):

yes

OpenStudy (ikram002p):

so sin 60 = Contrast /hypotenuse=B/C right ?

OpenStudy (anonymous):

yes

OpenStudy (ikram002p):

could u find cos 60 and tan 60 as well ?

OpenStudy (anonymous):

i believe cos 60 = c/a?

OpenStudy (anonymous):

and tan60 = a/b

OpenStudy (ikram002p):

well cos theta =adjacent /hypotenuse = A/C

OpenStudy (ikram002p):

cos 60 =A/C

OpenStudy (ikram002p):

and tan 60 = adjacent /Contrast

OpenStudy (ikram002p):

so tan 60 =B/A ok ?

OpenStudy (anonymous):

oh okay, so, do you mind helping more, like what are the sides?

OpenStudy (anonymous):

|dw:1407111090155:dw|

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