Question 1203332
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I'll use the greek letter {{{theta}}} (pronounced "theta") in place of t because theta is often used in trig.


sin(theta) = 2/3
sine = opposite/hypotenuse


We could say
opposite = 2
hypotenuse = 3


Draw a right triangle in quadrant Q2 (northwest quadrant)
The reference angle theta is placed in the bottom right corner of the triangle.
{{{drawing(400,400,-5,5,-5,5,
line(-5-3,0,5+3,0),
line(0,-5-3,0,5+3),
red(line(0,0,-3,4)),
red(line(-3,4,-3,0)),
red(line(-3,0,0,0)),
locate(-1,0.5,theta),

locate(-4,2,"Opp"),
locate(-2,-0.1,"Adj"),
locate(-1.5,2.4,"Hyp"),

locate(-4,1.5,"2"),
locate(-2,-0.6,"x"),
locate(-1,1.9,"3"),

locate(5-0.4,5-0.4,"Q1"),
locate(-4.6,5-0.4,"Q2"),
locate(-4.6,-4.6,"Q3"),
locate(5-0.4,-4.6,"Q4"),
locate(0.3,-4,matrix(1,2,"Diagram","not")),
locate(0.3,-4.4,matrix(1,2,"to","scale"))
)}}}
The unknown side adjacent is labeled x.


Use the pythagorean theorem
{{{a^2+b^2 = c^2}}}


{{{2^2+x^2 = 3^2}}}


{{{4+x^2 = 9}}}


{{{x^2 = 9-4}}}


{{{x^2 = 5}}}


{{{x = sqrt(5)}}} or {{{x = -sqrt(5)}}}
I'll go with the negative version of x to indicate we're to the left of the y axis.
x < 0 in Q2



Here's the updated diagram
{{{drawing(400,400,-5,5,-5,5,
line(-5-3,0,5+3,0),
line(0,-5-3,0,5+3),
red(line(0,0,-3,4)),
red(line(-3,4,-3,0)),
red(line(-3,0,0,0)),
locate(-1,0.5,theta),

locate(-4,2,"Opp"),
locate(-2,-0.1,"Adj"),
locate(-1.5,2.4,"Hyp"),

locate(-4,1.5,"2"),
locate(-2,-0.6,-sqrt(5)),
locate(-1,1.9,"3"),

locate(5-0.4,5-0.4,"Q1"),
locate(-4.6,5-0.4,"Q2"),
locate(-4.6,-4.6,"Q3"),
locate(5-0.4,-4.6,"Q4"),
locate(0.3,-4,matrix(1,2,"Diagram","not")),
locate(0.3,-4.4,matrix(1,2,"to","scale"))
)}}}
opposite = 2
adjacent = {{{-sqrt(5)}}}
hypotenuse = 3


Then,
{{{cos(theta) = adjacent/hypotenuse = -sqrt(5)/3}}}


{{{tan(theta) = opposite/adjacent = 2/(-sqrt(5)) = -2sqrt(5)/5}}}


{{{csc(theta) = hypotenuse/opposite = 3/2}}} (this is the reciprocal of sine)


{{{sec(theta) = hypotenuse/adjacent = 3/(-sqrt(5)) = -3sqrt(5)/5}}} (this is the reciprocal of cosine)


{{{cot(theta) = adjacent/opposite = -sqrt(5)/2}}} (this is the reciprocal of tangent)
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