Answer:
The trigonometric property that we will be using is :
{eq}sin^{2}(\theta )+cos^{2}(\theta )=1\\ \text{Lets proove this first}\\ \text{We know } sin(\theta) = \frac{perpendicular}{hypotenuse}\\ \\\text{We know } cos(\theta) = \frac{base}{hypotenuse}\\ \\So,\: sin^{2}(\theta )+cos^{2}(\theta )= (\frac{perpendicular}{hypotenuse})^{2} + (\frac{base}{hypotenuse})^2\\ \\\implies sin^{2}(\theta )+cos^{2}(\theta )= \frac{perpendicular^{2}+base^{2}}{hypotenuse^{2}}\\ \\\text{(According to pythgoras theorem } perpendicular^{2}+base^{2} = hypotenuse^{2})\\ \\\implies sin^{2}(\theta )+cos^{2}(\theta )= \frac{hypotenuse^{2}}{hypotenuse^{2}}\\ \\\implies sin^{2}(\theta )+cos^{2}(\theta )= 1\\ \\\text{(In our question }\theta = 50)\\ \\\\sin^{2}(50^{\circ} )+cos^{2}(50^{\circ} )=\:1 {/eq}