How do you find the opposite angle in a triangle when you only know the cosine angle degrees (30 degrees) and the hypotenuse length (20m)???
+9665 \(cos30 = \dfrac{adjacent \space side}{20}\)
\(adjacent \space side = 20 cos30 = 10\sqrt3\)
\(opposite \space side=\sqrt{20^2-(10\sqrt3)^2}=\sqrt{100}=10\)
\(atan(\dfrac{10\sqrt3}{10})= atan(\sqrt3)=60 \space degrees\)
Wait, if that is for right-angled triangles why I don't just 90 - 30 ...... never mind, at least I got the answer XD
+9665 \(cos30 = \dfrac{adjacent \space side}{20}\)
\(adjacent \space side = 20 cos30 = 10\sqrt3\)
\(opposite \space side=\sqrt{20^2-(10\sqrt3)^2}=\sqrt{100}=10\)
\(atan(\dfrac{10\sqrt3}{10})= atan(\sqrt3)=60 \space degrees\)
Wait, if that is for right-angled triangles why I don't just 90 - 30 ...... never mind, at least I got the answer XD
