# 1.4 Right triangle trigonometry  (Page 2/12)

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## Relating angles and their functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in [link] . The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

1. If needed, draw the right triangle and label the angle provided.
2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
3. Find the required function:
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

## Evaluating trigonometric functions of angles not in standard position

Using the triangle shown in [link] , evaluate $\mathrm{sin}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{cos}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{tan}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{sec}\text{\hspace{0.17em}}\alpha ,$ $\mathrm{csc}\text{\hspace{0.17em}}\alpha ,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\alpha .$

Using the triangle shown in [link] , evaluate and

## Finding trigonometric functions of special angles using side lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of $\text{\hspace{0.17em}}30°,$ $60°,$ and $\text{\hspace{0.17em}}45°,$ however, remember that when dealing with right triangles, we are limited to angles between

Suppose we have a $\text{\hspace{0.17em}}30°,60°,90°\text{\hspace{0.17em}}$ triangle, which can also be described as a triangle. The sides have lengths in the relation $\text{\hspace{0.17em}}s,\sqrt{3}s,2s.\text{\hspace{0.17em}}$ The sides of a $\text{\hspace{0.17em}}45°,45°,90°$ triangle, which can also be described as a $\text{\hspace{0.17em}}\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\text{\hspace{0.17em}}$ triangle, have lengths in the relation $\text{\hspace{0.17em}}s,s,\sqrt{2}s.\text{\hspace{0.17em}}$ These relations are shown in [link] .

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

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I know this work
salma
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Abhi
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20/(×-6^2)
Salomon
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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Abhi
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Abhi
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Abhi
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salma
Commplementary angles
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Cied
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Porter
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Porter
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Cesar
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Stotaw
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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