First
angle in front of c is 30°, so hypotenuse is equal to 2c
it's right triangle, so let's use Pythagorean theorem\
[tex](2c)^2=c^2+4^2\\4c^2=c^2+16\\3c^2=16\\\\c^2=\dfrac{16}{3} \\\\c=\dfrac{4}{\sqrt{3} } =\dfrac{4\sqrt{3} }{3 }[/tex]
Second
b - the base of the triangle, angles at the base are equal ⇒ cathets are equal
it's right triangle, so let's use Pythagorean theorem
[tex]b^2=7^2+7^2\\b^2=49+49\\b=\sqrt{2\times49} =7\sqrt{2}[/tex]
