how to find angle from trigonometric ratio without calculator?
Answers
Until now, we have used the calculator to evaluate the sine, cosine, and tangent of an angle. However, it is possible to evaluate the trig functions for certain angles without using a calculator.
This is because there are two special triangles whose side ratios we know! These two triangles are the 45-45-90 triangle and the 30-60-90 triangle.
The special triangles
30-60-90 triangles
A 30-60-90 triangle is a right triangle with a 30^\circ30
∘
30, degree degree angle and a 60^\circ60
∘
60, degree degree angle.
I'm skeptical. Can you show me how to derive these ratios?
45-45-90 triangles
A 45-45-90 triangle is a right triangle with two 45^\circ45
∘
45, degree degree angles.
I'm skeptical. Can you show me how to derive these ratios?
The trigonometric ratios of 30^\circ30
∘
30, degree
We are now ready to evaluate the trig functions of these special angles. Let's start with 30^\circ30
∘
30, degree.
Study the worked example below to see how this is done.
What is \sin(30^\circ)sin(30
∘
)sine, left parenthesis, 30, degree, right parenthesis?
Here's a worked example:
Step 1: Draw the special triangle that includes the angle of interest. Why?
30^\circ
30, degree30^\circ
30, degree303060609090
Step 2: Label the sides of the triangle according to the ratios of that special triangle.
Step 3: Use the definition of the trigonometric ratios to find the value of the indicated expression.
\begin{aligned} \sin (30^\circ) &= \dfrac{\text{opposite }}{\text{hypotenuse}} \\\\ &= \dfrac{x}{2x} \\\\ &= \dfrac{1\maroonD{\cancel{x}}}{2\maroonD{\cancel{x}}} \\\\ &=\dfrac{1}{2}\end{aligned}
sin(30
∘
)
=
hypotenuse
opposite
=
2x
x
=
2
x
1
x
=
2
1
Note that you can think of xxx as 1 x1x1, x so that it is clear that \dfrac{x}{2x}=\dfrac{1x}{2x}=\dfrac12
2x
x
=
2x
1x
=
2
1
start fraction, x, divided by, 2, x, end fraction, equals, start fraction, 1, x, divided by, 2, x, end fraction, equals, start fraction, 1, divided by, 2, end fraction.
Now let's use this method to find \cos(30^\circ)cos(30
∘
)cosine, left parenthesis, 30, degree, right parenthesis and \tan(30^\circ)tan(30
∘
)tangent, left parenthesis, 30, degree, right pa
renthesi
The trigonometric ratios of 45^\circ45
∘
45, degree
Let's try this process again with 45^\circ45
∘
45, degree. Here we can start by drawing and labeling the sides of a 45-45-90 triangle.
The trigonometric ratios of 60^\circ
∘
degree
The process of deriving the trigonometric ratios for the special angles 30^\circ30
∘
30, degree, 45^\circ45
∘
45, degree, and 60^\circ60
∘
60, degree is the same.
While we have not yet explicitly shown how to find the trigonometric ratios of 60^\circ60
∘
60, degree, we have all of the information we need!
A summary
We have calculated the trig ratios for 30^\circ30
∘
30, degree, 45^\circ45
∘
45, degree, and 60^\circ60
∘
60, degree. The table below summarizes our results.
\cos(\theta)cos(θ)cosine, left parenthesis, theta, right parenthesis \sin (\theta)sin(θ)sine, left parenthesis, theta, right parenthesis \tan( \theta)tan(θ)tangent, left parenthesis, theta, right parenthesis
\theta =30^\circθ=30
∘
theta, equals, 30, degree \greenD{\dfrac{\sqrt{3}}{2}}
2
3
start color greenD, start fraction, square root of, 3, end square root, divided by, 2, end fraction, end color greenD \greenD{\dfrac12}
2
1
start color greenD, start fraction, 1, divided by, 2, end fraction, end color greenD \greenD{\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}}}
3
3
=
3
1
start color greenD, start fraction, square root of, 3, end square root, divided by, 3, end fraction, equals, start fraction, 1, divided by, square root of, 3, end square root, end fraction, end color greenD
\theta = 45^\circθ=45
∘
theta, equals, 45, degree \purpleC{\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}}
2
2
=
2
1
start color purpleC, start fraction, square root of, 2, end square root, divided by, 2, end fraction, equals, start fraction, 1, divided by, square root of, 2, end square root, end fraction, end color purpleC \purpleC{\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}}
2
2
=
2
1
start color purpleC, start fraction, square root of, 2, end square root, divided by, 2, end fraction, equals, start fraction, 1, divided by, square root of, 2, end square root, end fraction, end color purpleC \purpleC11start color purpleC, 1, end color purpleC
\theta = 60^\circθ=60
∘
theta, equals, 60, degree \greenD{\dfrac12}
2
1
start color greenD, start fraction, 1, divided by, 2, end fraction, end color greenD \greenD{\dfrac{\sqrt{3}}{2}}
2
3
start color greenD, start fraction, square root of, 3, end square root, divided by, 2, end fraction, end color greenD \greenD{\sqrt{3}}
3
start color greenD, square root of, 3, end square root, end color greenD
These values tend to occur often in advanced trigonometry problems. Because of this, it is helpful to know them.
Some people choose to memorize these values, but memorization is not necessary. In this article, you derived the values yourself, so hopefully you can re-derive them whenever you need them in the Future
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