# How do I solve this question?

May 2, 2018

Let's assume a right-angled triangle ABC with base AB = $5 x$ and hypotenuse AC = $7 x$.

By Pythagoras theorem, we have: $B {C}^{2} = A {C}^{2} - A {B}^{2}$
BC is the perpendicular.
By definition, sin(t) is the ratio of the perpendicular to the hypotenuse of a right-angled triangle.
$\sin t = \frac{\sqrt{A {C}^{2} - A {B}^{2}}}{A C}$
$\implies \sin \left(t\right) = \frac{\sqrt{49 {x}^{2} - 25 {x}^{2}}}{7 x}$

Since the sine of any angle is a constant, irrespective of side lengths, we may assume $x$ to be any number we wish. Let's assume it to be 1.
$\implies \sin t = \frac{\sqrt{24}}{7} = \frac{2 \sqrt{6}}{7}$

(Note, we could have used the identity ${\sin}^{2} x + {\cos}^{2} x = 1$ too)

The function cos(t) is symmetric about the y-axis. This means cos(-t) = cos(t)
$\implies \cos \left(- t\right) = - \frac{5}{7}$