What is the Method to Calculate Sin β in a Triangle?

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In summary, the numerical value of sin β for the triangle shown is 2√13/13, and it can be found using the Pythagorean theorem or by using a calculator to find the value of atan(2/3) and then using the sine function.
  • #1
basty
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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?

triangle.png
 
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  • #2
basty said:
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?

triangle.png

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
You can use the pythagorean theorem in triangle ABD to find y with respect to x and then find sin beta
 
  • #4
Mastermind01 said:
You can use the pythagorean theorem in triangle ABD to find y with respect to x and then find sin beta

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
basty said:
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?

That is correct.

You can even use a calculator to tally.

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

sin(33) = 0.546 = 2 / sqrt(13)
 

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