Question

Cos 43° = x/√x²+y² then tan 47°

Answer 1

Answer:

$\frac{x}{y}$

Step-by-step explanation:

Given,

$\cos 43^{\circ}=\frac{x}{\sqrt{x^2 + y^2}}$

$\implies \cos (90^{\circ}-47^{\circ}) = \frac{x}{\sqrt{x^2 + y^2}}$

$\implies \sin 47^{\circ}= \frac{x}{\sqrt{x^2 + y^2}}$

Now,

$\sin \theta = \frac{P}{H}$

By the Pythagoras theorem,

$H^2 = P^2 + B^2$

$\implies B=\sqrt{H^2 - P^2}=\sqrt{x^2 + y^2 - x^2}=\sqrt{y^2}=y$

Thus,

$\tan 47^{\circ}=\frac{P}{B}=\frac{x}{y}$