Trigonometric Ratios and the Pythagorean Theorem
Task
 In the triangle pictured above show that $$ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1 $$
 Deduce that $\sin^2{\theta} + \cos^2{\theta} = 1$ for any acute angle $\theta$.
 If $\theta$ is in the second quadrant and $\sin{\theta} = \frac{8}{17}$ what can you say about $\cos{\theta}$? Draw a picture and explain.
IM Commentary
The purpose of this task is to use the Pythagorean Theorem to establish the fundamental trigonometric identity $\sin^2{\theta} + \cos^2{\theta} = 1$ for an acute angle $\theta$. The reasoning behind this identity is then applied to calculate $\cos{\theta}$ for a given obtuse angle. In order to successfully complete part (c) students must be familiar with the definitions of trigonometric functions for arbitrary angles using the unit circle (FTF.2).
The Pythagorean Theorem requires $\triangle ABC$ to be a right triangle and so $\angle A$ must be acute. So the reasoning in parts a and b of this task establish the identity $$ \sin^2{\theta} + \cos^2{\theta} = 1 $$ for acute angles only. Using the unit circle as in part c, however, shows that this identity is true for all angles. More about this can be found in http://www.illustrativemathematics.org/tasks/1692.
Solution

The Pythagorean Theorem says that if $\triangle ABC$ is a right triangle with right angle $B$ then $AB^2 + BC^2 = AC^2$. Dividing both sides by $AC^2$ gives $$ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1. $$

If $0 \lt m(\theta) \lt 90$, then we can make a right triangle $ABC$, as pictured in the problem statement, so that $m(\angle BAC) = \theta$:
Then from part (a) we have $$ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1. $$ We also know that $\frac{AB}{AC} = \cos{\theta}$ and $\frac{BC}{AC} = \sin{\theta}$ so we have $\sin^2{\theta} + \cos^2{\theta} = 1$.

Below is a picture of an angle $\theta$ in the second quadrant with $\sin{\theta} = \frac{8}{17}$:
In the picture, the purple circle is the unit circle. The coordinates of $C$ are $(\cos{\theta}, \sin{\theta})$ and since $C$ lies on the unit circle we have $$ \sin^2{\theta} + \cos^2{\theta} = 1. $$ Since $\sin{\theta} = \frac{8}{17}$ we can solve for $\cos{\theta}$ and we find $\cos{\theta} = \pm \frac{15}{17}$. Since we are in the second quadrant $\cos{\theta} =  \frac{15}{17}$.
Trigonometric Ratios and the Pythagorean Theorem
 In the triangle pictured above show that $$ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1 $$
 Deduce that $\sin^2{\theta} + \cos^2{\theta} = 1$ for any acute angle $\theta$.
 If $\theta$ is in the second quadrant and $\sin{\theta} = \frac{8}{17}$ what can you say about $\cos{\theta}$? Draw a picture and explain.