+118653 if sin(t)=5/7, and cos<0, then what is cos(t)?
If sin is positive and cos is negative then the angle must be in the 2nd quadrant
actually that doesn't even need to be thought about.
Draw a right angled triangel and put ß as one of the acute angles. (I won't use t because t is an obtuse angle)
The opposite side is 5 the hypotenuse is 7
So the adjacent is $$\sqrt{7^2-5^2}=\sqrt{49-25}=\sqrt{24}$$
$$Cos(t) = -\frac{adj}{opp} = -\frac{\sqrt{24}}{7}$$
(I added the minus because it is the obtuse angle equivalent.)
+118653 if sin(t)=5/7, and cos<0, then what is cos(t)?
If sin is positive and cos is negative then the angle must be in the 2nd quadrant
actually that doesn't even need to be thought about.
Draw a right angled triangel and put ß as one of the acute angles. (I won't use t because t is an obtuse angle)
The opposite side is 5 the hypotenuse is 7
So the adjacent is $$\sqrt{7^2-5^2}=\sqrt{49-25}=\sqrt{24}$$
$$Cos(t) = -\frac{adj}{opp} = -\frac{\sqrt{24}}{7}$$
(I added the minus because it is the obtuse angle equivalent.)
