First Part 45-45-90 Triangle
first triangle
where the two angles different to 90° are same, the measure of the legsof the triangle are the same
then
[tex]A=5[/tex]
and to calculate B or the hypotenuse we use pythagoras
[tex]a^2+b^2=h^2[/tex]
where a and b are legs and h the hypotenuse
replacing
[tex]\begin{gathered} 5^2+5^2=h^2 \\ 25+25=h^2 \\ 50=h^2 \\ h=\sqrt[]{50} \\ h=5\sqrt[]{2} \end{gathered}[/tex]
the hypotenuse or B is
[tex]B=5\sqrt[]{2}[/tex]
Second triangle
legs of the triangle have the same value then if we apply pythagoras
[tex]a^2+b^2=h^2[/tex]
and replace the legs with the same value(a)
[tex]\begin{gathered} a^2+a^2=h^2 \\ 2a^2=h^2 \end{gathered}[/tex]
we can replace the hypotenuse and solve a
[tex]\begin{gathered} 2a^2=(3\sqrt[]{2})^2 \\ 2a^2=18 \\ a^2=\frac{18}{2} \\ \\ a=\sqrt[]{9} \\ a=3 \end{gathered}[/tex]
value of each leg is 3 units, then
[tex]C=D=3[/tex]
Second part 30-60-90 triangle
First triangle
we use trigonometric ratios to solve, for example I can use tangent for the angle 60 to find E
[tex]\tan (\alpha)=\frac{O}{A}[/tex]
where alpha is the angle, O the oppiste side of the angle and A the adjacet side of the angle
using angle 60°
[tex]\begin{gathered} \tan (60)=\frac{E}{6} \\ \\ E=6\tan (60) \\ \\ E=6\sqrt[]{3} \end{gathered}[/tex]
now using sine we calculate F or the hypotenuse
[tex]\sin (\alpha)=\frac{O}{H}[/tex]
where alpha is the angle, O the opposite side from the angle and H the hypotenuse
using angle 60°
[tex]\begin{gathered} \sin (60)=\frac{E}{F} \\ \\ F=\frac{E}{\sin (60)} \\ \\ F=\frac{6\sqrt[]{3}}{\sin (60)} \\ \\ F=12 \end{gathered}[/tex]
Second triangle
we use sine with 60° to find H
[tex]\begin{gathered} \sin (\alpha)=\frac{O}{h} \\ \\ \sin (60)=\frac{H}{20} \\ \\ H=20\sin (60) \\ H=10\sqrt[]{3} \end{gathered}[/tex]
use cosine with 60° to find G
[tex]\begin{gathered} \cos (\alpha)=\frac{A}{h} \\ \\ \cos (60)=\frac{G}{20} \\ \\ G=20\cos (60) \\ \\ G=10 \end{gathered}[/tex]
Final Values
[tex]\begin{gathered} A=5 \\ B=5\sqrt[]{2} \\ C=3 \\ D=3 \\ E=6\sqrt[]{3} \\ F=12 \\ G=10 \\ H=10\sqrt[]{3} \end{gathered}[/tex]
