1. Since we have 2 equal angles in our triangle, we will also have two equal sides. Given that the third side is fixed at 10 inches, we can only form two triangle with those measures.
2. Check the attached picture.
3. First, we are going to use the interior angle theorem to find the third angle of our triangle. Remember that the interior angle theorem says that the sum of the interior angles of a triangle is always 180°. Lets apply it to our triangle: we know that the measure of two of its angles is 40°; let A be our third angle:
[tex]A+40+40=180[/tex]
[tex]A+80=180[/tex]
[tex]A=100[/tex]
Now that we have the three angles of our triangle, we can use the law of sines to find the sides of our first triangle:
[tex] \frac{b}{sin40} = \frac{10}{sin100} [/tex]
[tex]b= \frac{10sin40}{sin100} [/tex]
[tex]b=6.5[/tex]
[tex] \frac{c}{sin40} = \frac{6.5}{sin40} [/tex]
[tex]c= \frac{6.5sin40}{sin40} [/tex]
[tex]c=6.5[/tex]
Now, lets do the same for our second triangle:
[tex] \frac{a}{sin100} = \frac{10}{sin40} [/tex]
[tex]a= \frac{10sin100}{sin40} [/tex]
[tex]a=15.3[/tex]
[tex] \frac{c}{sin40} = \frac{15.3}{sin40} [/tex]
[tex]c= \frac{15.3sin40}{sin40} [/tex]
[tex]c=15.3[/tex]
We can conclude that we used the law of sines because the problem gave us tow angles and one side.
