Trigonometry

how is sin a = perpendicular by base

V Shalini

11 Years agoGrade

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

First what you have written is wrong.
$sina = \frac{Perpendicular}{Hypotenuse}$
$tana = \frac{Perpendicular}{Base}$
Well there is no particular reason behind this.
You can understand in this way.
tan(a) is defined as a slope.
Lets assume a point P(x, y) in the coordinate plane.
Join this point to origin O(0, 0).
Now, slope of OP is
$\frac{y}{x}$
Projection of OP on x-axis is x. Lets point A (x, 0)
Now we have triangle OAP, in which
$tan(a) = \frac{y}{x} = \frac{AP}{OA}$
$tan(a) = \frac{Perpendicular}{Base}$
By Definition, we have
$tan(a) = \frac{sin(a)}{cos(a)}$
From here, we have
$sin(a) = \frac{Perpendicular}{Hypotenuse}$