# 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?

• I didnt check with calculator, but in theory your approach seems correct to me. – Paolo Leonetti Feb 11 '15 at 23:09
• You cannot have two obtuse angles in the same triangle. Keep in mind that $\sin\theta = \sin(180^\circ - \theta)$. – N. F. Taussig Feb 11 '15 at 23:41

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 $$m\angle CBD + m\angle BCD + m\angle BDC = 7.8^\circ + 10.5^\circ + 161.7^\circ = 180^\circ$$

Hint: $\sin(B)=0.4811$ means the reference angle is $28.8^\circ$. Thus $B=28.8^\circ$ or $B=151.2^\circ$.

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 $$u = 42.258, b = 23.7411, y = \cos^{-1}(b/25) = 18.26^\circ, x = 161.74^\circ$$