Answer:
[tex]\frac{25\sqrt{3} }{8}[/tex]
Step-by-step explanation:
CosA = [tex]\frac{1}{2}[/tex] , so <A is 60º
Hypotenuse (c) is 5
Opposite is (b)
Adjascent is (a)
cos A = [tex]\frac{adjascent}{hypotenuse}[/tex] -----> [tex]\frac{1}{2} = \frac{a}{5}[/tex] ---------> a = [tex]\frac{5}{2}[/tex]
Pytagoras
[tex]5^{2} = (\frac{5}{2}) ^{2} + b^{2}\\ b^{2} = 25 - \frac{25}{4} \\b^{2} = \frac{75}{4} \\\\b = \frac{5\sqrt{3} }{2}[/tex]
A = b . a . 1/2
A = [tex]\frac{25\sqrt{3} }{8}[/tex]
CosA = , so <A is 60º
Hypotenuse (c) is 5
Opposite is (b)
Adjascent is (a)
cos A = -----> ---------> a =
Pytagoras
A = b . a . 1/2
A =
Step-by-step explanation:
