Question

If 0°<0<90° and sin0=12/13, find cos0 using trigonometric identities.

Answer 1

here's the solution,

we know :

$\sin( \theta) = \dfrac{perpendicular}{hypotenuse}$

So,

$\dfrac{p}{h} = \dfrac{12}{13}$

so.. let the perpendicular be 12x and hypotenuse be 13x

now,

by applying pythagoras theorem,

• $b {}^{2} = h {}^{2} - p {}^{2}$

where,

• b = base
• h = hypotenuse
• p = perpendicular

So,

• $b {}^{2} = ({13x})^{2} - ({12x})^{2}$
• $b {}^{2} = 169x {}^{2} - 144x {}^{2}$
• ${b}^{2} = 25x {}^{2}$
• $b = 5x$

so,

• $\cos( \theta) = \dfrac{b}{h}$

• $\cos( \theta) = \dfrac{5x}{13x}$

• $\cos( \theta) = \dfrac{5}{13}$

hope it helps !!