The problem gives us information about our cosine and our sine. Cosine is 6 over radical 61 and our sine is greater than 1. With our knowledge of ASTC we can say that cosine is positive in quadrants one and four and that sine is positive in quadrants one and two. Our triangle should be in quadrant one, even though we put the triangle in quadrant two, we messed up on the very first step. Since we have cosine of theta we can use the reciprocal identity of sec of theta=1/cos of theta. Our cosine is 6 radical 61 and to cancel out the denominator of 6 radical 61. The problem ends up being 1/6 radical 61 but we multiply by the reciprocal and secant should equal 6 radical 61/61. Next we solved for sin used the pythagorean identity of 1-cos^2theta=sin^2theta. 6/radical 61 was used for cos^2theta, it ended up becoming 36/61, next we subtracted it by 1 and ended up with radical 25/61, then we get 5/radical 61, finally we multiply the numerator and denominator by radical 61 and sin theta=5radical61/61. Since we have cosine and sin we used the ration identity of cot theta=cos theta/sin theta. With this ratio set up radical 61 over 61 canceled out in the numerator and the denominator and cot theta ends up equaling 6/5. Finally we use the reciprocal identity of tan theta=1/cot theta. The function should be 1/(6/5) and we multiply by the reciprocal and tan theta equals 5/6.
We must now use SOHCAHTOA and we should start off by using the pythagorean theorem to find the hypotenuse. The hypotenuse ends up being radical 61 and now we find our trig functions. The ratio for sin theta is y/r and the ratio for cosecant theta is r/y. Sin theta is 5radical61/61 and cosecant theta is radical 61/5. The ratio for cosine theta is x/r and the ratio for secant is r/x. Cosine theta is -6radical61/61 and secant is -radical 61/6. The ratio for tangent theta is y/x and the ratio for cotangent is x/y. Tangent is -5/6 and cotangent is -6/5.
