Given the following sides of the triangle:
a = 5, b = 5, c = 5 √2
Let's find the cos of angle F:
[tex]\cos ^{-1}(a^2+b^2-c^2)\frac{2ab}{\square}[/tex][tex]\cos ^{-1}(5^2\text{ +( 5}\sqrt{2})^2\text{ - }5^2)\frac{\square}{\square}2\cdot\text{ 5 }\cdot\text{ 5}\sqrt{2}[/tex][tex]F\text{ }\approx\text{ 45}\circ\text{ }\Rightarrow\text{ cos 45 = (}\sqrt{(2})\frac{2}{\square}[/tex]
The correct option is D. Cos (D) because the sides a and b are equal and this is an isosceles triangle
In addition, angles D and F are equal (45 degrees)
