Consider the upper triangle
if we want to find b side, consider the following trigonometric identity:
[tex]\cos \text{ (}\theta\text{) = }\frac{adjacent\text{ side}}{hypotenuse}[/tex]
in this case, we have:
[tex]\cos \text{ (45) = }\frac{b}{20}[/tex]
solve for b:
[tex]b\text{ = cos(45) x 20 = }\frac{\sqrt[]{2}}{2}\text{ x 20 = 10}\sqrt[]{2}[/tex]
then, we can conclude that
[tex]b\text{ = 10}\sqrt[]{2}[/tex]
Now, for a-side, consider the following trigonometric identity:
[tex]\sin \text{ (}\theta\text{) = }\frac{opposite\text{ side}}{hypotenuse}[/tex]
in this case, we have:
[tex]\sin \text{ (}\theta\text{) = }\frac{a}{20}[/tex]
solve for a:
[tex]a\text{ = }\sin \text{(45) x 20 = }\frac{\sqrt[]{2}}{2}\text{ x 20 = 10}\sqrt[]{2}[/tex]
then, we can conclude that
[tex]a\text{ = 10}\sqrt[]{2}[/tex]
Now, for the c-side, consider the greater triangle :
if we denote the hypotenuse by h, then by Pythagorean theorem we have:
[tex]h^2=20^2+15^2[/tex]
but h = a + c, then, replacing this in the previous equation we have
[tex]h^2=(a+c)^2=20^2+15^2[/tex]
but, we know that a is
[tex]a\text{ = 10}\sqrt[]{2}[/tex]
then we have:
[tex](10\sqrt[]{2}+c)^2=20^2+15^2\text{ = 625}[/tex]
now, taking the square root of both sides of the equation we obtain:
[tex]10\sqrt[]{2}+c^{}=\text{ 25}[/tex]
solve for c:
[tex]c^{}=\text{ 25}-\text{ 10}\sqrt[]{2\text{ }}\text{ }\approx10.85[/tex]
then we can conclude that :
[tex]a\text{ = 10}\sqrt[]{2}[/tex]
[tex]b\text{ = 10}\sqrt[]{2}[/tex]
and
[tex]c^{}=\text{ 25}-\text{ 10}\sqrt[]{2\text{ }}\text{ }\approx10.85[/tex]
