I think pencil and paper would be the best method
![\bf \begin{array}{clclll} -6&+&6\sqrt{3}\ i\\ \uparrow &&\uparrow \\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right) \end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bclclll%7D%0A-6%26%2B%266%5Csqrt%7B3%7D%5C%20i%5C%5C%0A%5Cuparrow%20%26%26%5Cuparrow%20%5C%5C%0Aa%26%26b%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5C%0A%5Ctheta%20%3Dtan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7Bb%7D%7Ba%7D%20%5Cright%29%0A%5Cend%7Bcases%7D%5Cqquad%20r%5Bcos%28%5Ctheta%20%29%2Bi%5C%20sin%28%5Ctheta%20%29%5D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.
thus
![\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]](https://tex.z-dn.net/?f=%5Cbf%2012%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%2Bi%5C%20sin%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%5Cright%5D)
OK, let's try with no figure. We have an isosceles triangle sides s,s, and b.
Opposite b is angle t.
Draw the altitude h to bisect t. We have two right triangles, legs b/2 and h, hypotenuse s. The angle opposite b/2 is t/2 so
sin(t/2) = (b/2)/s = b/2s
So we arrived at the first part,
b = 2s sin(t/2)
The area of a triangle with sides s,s and included angle t is
A = (1/2) s² sin t
Answer:
Step-by-step explanation:
(5/6) itself is less than 1 (but greater than 0). Thus, anything multiplied by (5/6) will result in a smaller-by-magnitude quantity.
For example, (5/6)(3/4) is LESS than 3/4 because 5/6 is greater than zero but less than 1.
