Sine Law Homework Question I am given a triangle with two triangles inside of it, and am asked to solve for angles $x$ and $y$. 
I have illustrated the triangle here:

My process: 
$$\frac{25}{\sin10.5} = \frac{66}{\sin B}$$
We then solve for angle $B$ (the top one) and we get $28.8$ degrees.
$$\begin{align}
28.8 + 10.5 & = 39.3 \\
180 - 39.3 & = 140.7
\end{align}$$
Therefore, the angle on the left, $y$, is equal to $140.7$. Then, by the property of isosceles triangles, the one on the right is also $140.7$. $180 - 140.7 = 39.3$, giving us angle $X$.
$$\begin{align}
\angle X & = 39.3^\circ \\
\angle Y & = 140.7^\circ
\end{align}$$
The textbook answers shows that this is wrong. Am I wrong, or is the textbook?
 A: When you solved the Law of Sines, you should have obtained two solutions for $m\angle ABC$.  The one you chose does not produce a triangle, so you have to use the other solution.
Consider the diagram below.

By the Law of Sines, 
$$\frac{25}{\sin(10.5^\circ)} = \frac{66}{\sin\theta}$$
Solving for $\sin(\theta)$ yields
$$\sin\theta = \frac{66\sin(10.5^\circ)}{25}$$
One solution of the equation is 
$$\theta = \arcsin\left[\frac{66\sin(10.5^\circ)}{25}\right] = 28.8^\circ$$
where the answer has been rounded to one decimal place.  Since $\sin\alpha = 180^\circ - \sin\alpha$, the other solution is 
$$\theta = 180^\circ - \arcsin\left[\frac{66\sin(10.5^\circ)}{25}\right] = 151.2^\circ$$
where the answer has been rounded to one decimal place.
If $\theta = 28.8^\circ$, then since the sum of the measures of $\triangle ABC$ is $180^\circ$,
$$y = 180^\circ - 10.5^\circ - 28.8^\circ = 140.7^\circ$$
We were given that $\overline{AB} \cong \overline{DB}$, so $\triangle ABD$ is isosceles.  By the Isosceles Triangle Theorem, $\angle BAD \cong \angle BDA$.  Thus, 
$$m\angle BDA = m\angle BAD = y = 140.7^\circ$$
which is impossible since the Angle Sum Theorem for Triangles implies that a triangle cannot contain two obtuse angles.  Hence, we must discard the solution $\theta = 28.8^\circ$. 
If $\theta = 151.2^\circ$, then 
$$y = 180^\circ - 10.5^\circ - 151.2^\circ = 18.3^\circ$$
Thus, by the Isosceles Triangle Theorem, 
$$m\angle ADB = m\angle BAD = y = 18.3^\circ$$
Since the sum of the measures of $\triangle ABD$ is $180^\circ$, 
$$m\angle ABD = 180^\circ - m\angle BAD - m\angle ADB = 180^\circ - 2 \cdot 18.3^\circ = 143.4^\circ$$
Observe that $\angle BDC$ is an exterior angle of $\triangle ABD$.  By the Exterior Angle Theorem,
$$x = m\angle BDC = m\angle BAD + m\angle ABD = 18.3^\circ + 143.4^\circ = 161.7^\circ$$
As confirmation, observe that 
$$\varphi = m\angle CBD = m\angle ABC - m\angle ABD = 151.2^\circ - 143.4^\circ = 7.8^\circ$$
and that 
$$m\angle CBD + m\angle BCD + m\angle BDC = 7.8^\circ + 10.5^\circ + 161.7^\circ = 180^\circ$$
A: Hint: $\sin(B)=0.4811$ means the reference angle is $28.8^\circ$. Thus $B=28.8^\circ$ or $B=151.2^\circ$.
A: let the base of the isosceles triangle has base $2a$ and height $b.$ then we can get two relations between and $a$ and $b$ as follows:
$$a = (66-b)\tan 10.5^\circ,\, a^2 + b^2 = 25^2. $$
we can find $b$ by solving the quadratic equation $$ (u\tan 10.5^\circ)^2 + (66-u)^2 = 25^2$$ where $u = 66-b.$
the solution $u < 66$ is $$u = 42.258, b = 23.7411, y = \cos^{-1}(b/25) = 18.26^\circ, x = 161.74^\circ  $$
