# How To Find Sin β?

1. Oct 3, 2015

### basty

How do you find the numerical value sin β for the triangle shown on below image?

I can only find

$\tan β = \frac{AB}{BD} = \frac{2x}{3x} = \frac{2}{3} = 0.666666667$

then

$β = \tan^{-1} 0.666666667 = 0.59°$

then

$\sin β = \sin 0.59° = 0.0103$

Is there another method to find the numerical value of sin β?

2. Oct 3, 2015

### SteamKing

Staff Emeritus
First of all, tan-1(2/3) ≠ 0.59°

There are two common angular measures in use: degrees and radians. The calculators we use to compute the trig functions and their inverses need to be set on one measure or the other in order to perform the correct calculation.

The tangent of a 45° angle = 1, so the angle whose tangent is 2/3 will be closer to 45° than to 0°.

The Pythagorean Identity, sin2(θ) + cos2(θ) = 1, can be manipulated to give

tan2(θ) + 1 = sec2(θ) or
cot2(θ) + 1 = csc2(θ),

where
csc(θ) = 1 / sin(θ)
sec(θ) = 1 / cos(θ)
cot(θ) = 1 / tan(θ)

3. Oct 3, 2015

### Mastermind01

You can use the pythagorean theorem in triangle ABD to find y with respect to x and then find sin beta

4. Oct 3, 2015

### basty

From the pythagorean formula, I get:

$y^2 = (2x)^2 + (3x)^2$
$y^2 = 4x^2 + 9x^2$
$y^2 = 13x^2$
$y = \sqrt{13x^2}$
$y = \sqrt{13}x$

$\sin β = \frac{2x}{\sqrt{13}x} = \frac{2}{\sqrt{13}} = \frac{2}{\sqrt{13}} × \frac{\sqrt{13}}{\sqrt{13}} = \frac{2\sqrt{13}}{13}$

Is that correct?

5. Oct 3, 2015

### Mastermind01

That is correct.

You can even use a calculator to tally.

atan(2/3) = 33 degrees (approximately)

sin(33) = 0.546 = 2 / sqrt(13)