Answer:
Explanation:
Let us draw the triangle first.
Now, the law of cosines gives
[tex]10^2=8^2+x^2-2\cdot8\cdot x\cos (80^o)[/tex]simplifying the above gives
[tex]100=64+x^2-16x\cos (80^o)[/tex]subtracting 64 from both sides gives
[tex]100-64=x^2-16x\cos (80^o)[/tex][tex]\Rightarrow36=x^2-16x\cos (80^o)[/tex]Rewriting the above in a more familiar form gives
[tex]x^2-16\cos (80^o)x-36=0[/tex]which is a quadratic equation.
Using the quadratic formula we solve for x and get
[tex]x=\frac{16\cos (80^o)\pm\sqrt[]{(16\cos (80^o))^2-4(1)(-36)}}{2(1)}[/tex]whose positive solution is
[tex]\boxed{x=7.54791\ldots}[/tex]Next, we find the angle Θ.
To find Θ, we use the law of sines.
The law of sines in our case gives
[tex]\frac{10}{\sin(80^o)}=\frac{8}{\sin \theta}[/tex]cross multipication gives
[tex]10\sin \theta=8\sin (80^o)[/tex]dividing both sides by 10 gives
[tex]\sin \theta=\frac{8\sin (80^o)}{10}[/tex]With the help of a calculator, we evaluate the right-hand side and get
[tex]\sin \theta=0.787846[/tex]taking the inverse sine of both sides gives
[tex]\begin{gathered} \boxed{\theta=51.984688^o\ldots^{}} \\ \end{gathered}[/tex]Last but not least, we find the value of α.
To find α, we use the fact that the sum of internal angles of a triangle must be 180°.
Therefore, we have
[tex]\alpha+\theta+80^o=180^o[/tex]putting in the value of Θ gives
[tex]\alpha+51.984688^o+80^o=180^o[/tex]simplifying and solving for α gives
[tex]\boxed{\alpha=48.015^o}[/tex]Hence, our angle and side length measures rounded to the nearest tenth are:
[tex]\begin{gathered} x=7.6 \\ \theta=52.0^o \\ \alpha=48.0^o \end{gathered}[/tex]
