Trigonometry / Sum of two angles (α + β) if sinα = 8/17 and sinβ = 15/17 Find the sum of two angles α and β if sinα = 8/17 and sinβ = 15/17 if they are
A) acute
B) obtuse
How do you approach this problem? I'm stuck at the beginning. Please help.
 A: In any problem like this where you are given the $\sin$ of an angle as a fraction, there is usually some involvement of Pythagoras' Theorem.
Imagine a right-angled triangle with opposite side length 8 and hypotenuse length 17. It will have the angle $\alpha$ with $\sin \alpha = \frac 8 {17}$. Find the other side...
A: I hope this is a simpler method.
A) When the angles are acute
When $\sin\alpha = \frac{8}{17} \implies \cos\alpha = \frac{15}{17}$.
When $\sin\beta = \frac{15}{17} \implies \cos\beta = \frac{8}{17}$.
$$\sin{(\alpha + \beta)}= \sin\alpha\cos\beta + \sin\beta\cos\alpha$$
Substituting the values above we get,
$$\sin{(\alpha + \beta)} = 1$$
$$\therefore \alpha + \beta = \frac{\pi}{2}$$
B) When the angles are obtuse
$$\sin{(\alpha+\beta)} = 1$$
$$\alpha + \beta = n\pi + (-1)^{n} \frac{\pi}{2}, n\in\Bbb{Z}$$
The angles selected from the set of solutions here must be greater than $\frac{\pi}{2}$.
A: $$
\sin(\alpha) = \frac{8}{17}\\
\sin(\beta) = \frac{15}{17}
$$
The sine of each of these angles are positive.
$$\alpha, \beta \in [0,\pi]$$
If they are acute then they lie between $0$ and $\frac{\pi}{2}$.
If they are obtuse they lie between  $\frac{\pi}{2}$ and $\pi$
If the angles are obtuse, then they will be $\pi - \theta_a$ where $\theta_a$ is the acute angle.
$$ sin^{-1} \frac{8}{17} =~ .489$$
$$ sin^{-1} \frac{15}{17} =~ 1.080$$
$$\alpha_o =~ \pi - .489 =~ 2.65169265 $$
$$\beta_o =~ \pi - 1.080 =~ 2.061$$
