Two thirds of a number
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well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus
<span>Using: a^2 + b^2 = c^2 (pythagoras theorem)
where 'a' and 'b' are the legs of a right triangle, c is the hypotenuse
</span><span>compare the triangle I just drew to the triangle you have for your problem
</span><span>you'll see that a = unknown b = 18 c = 36
</span><span>a=18sqrt3
=31.2</span>
where 'a' and 'b' are the legs of a right triangle, c is the hypotenuse
</span><span>compare the triangle I just drew to the triangle you have for your problem
</span><span>you'll see that a = unknown b = 18 c = 36
</span><span>a=18sqrt3
=31.2</span>