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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.

Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha/opposite to beta, and adjacent to beta/opposite alpha.
The side adjacent to one angle is opposite the other.

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 sin α , cos α , tan α , sec α , csc α , and cot α .

Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled.
sin α = opposite  α hypotenuse = 4 5 cos α = adjacent to  α hypotenuse = 3 5 tan α = opposite  α adjacent to  α = 4 3 sec α = hypotenuse adjacent to  α = 5 3 csc α = hypotenuse opposite  α = 5 4 cot α = adjacent to  α opposite  α = 3 4

Using the triangle shown in [link] , evaluate sin   t , cos   t , tan   t , sec   t , csc   t , and cot   t .

Right triangle with sides 33, 56, and 65. Angle t is also labeled.

s i n   t = 33 65 , cos   t = 56 65 , t a n   t = 33 56 , sec   t = 65 56 , csc   t = 65 33 , cot   t = 56 33

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 30° , 60° , and 45° , however, remember that when dealing with right triangles, we are limited to angles between  and 90° .

Suppose we have a 30° , 60° , 9 triangle, which can also be described as a π 6 ,   π 3 , π 2 triangle. The sides have lengths in the relation s , 3 s , 2 s . The sides of a 45° , 45° , 90° triangle, which can also be described as a π 4 , π 4 , π 2 triangle, have lengths in the relation s , s , 2 s . These relations are shown in [link] .

Two side by side graphs of circles with inscribed angles. First circle has angle of pi/3 inscribed. Second circle has angle of pi/4 inscribed.
Side lengths of special triangles

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

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
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
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
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Uday
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salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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I'm interested in nanotube
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what is nanomaterials​ and their applications of sensors.
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what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
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Azam
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I'm interested in Nanotube
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Prasenjit Reply
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.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Essential precalculus, part 2. OpenStax CNX. Aug 20, 2015 Download for free at http://legacy.cnx.org/content/col11845/1.2
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