what are the ratios for sin a and cos a? the diagram is not drawn to scale.
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How to write sohcahtoa ratios, given side lengths
What do we hateful past 'ratio' of sides?
Sine, cosine and tangent of an angle represent the ratios that are e'er truthful for given angles. Recollect these ratios only apply to right triangles.
The 3 triangles pictured below illustrate this.
Diagram i
Although the side lengths are dissimilar , each one has a 37° angle, and every bit you tin see, the sine of 37 is ever the same!
In other words, $$sin(\carmine { 37^{\circ} } )$$ is ever $$ \red {.6 } $$ .
(Annotation: I rounded to the nearest 10th) That's, of course, why we can use a reckoner to detect these sine, cosine and tangent ratios.
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Practice Bug
Trouble i
Diagram 2
Step one
In the diagram 2, what side is adjacent to $$\angle L$$?
Step 2
What side is the hypotenuse?
Step 3
Calculate $$cos(\angle L)$$.
Use sochcahtoa to help remember the ratio.
$ cos(\bending \ruddy L) = \frac{adjacent}{hypot} \\ cos(\angle \ruby Fifty) = \frac{8}{x} \\ = .8 $
Footstep iv
Calculate $$ cos(\angle North)$$ (a different angle from prior question, look carefully at letters).
$ cos(\angle \red N) = \frac{6}{10} \\ = .half dozen $
Trouble 2
What is the sine ratio of $$\bending C$$ ?
Recall the sine ratio
$ sin(\angle C) =\frac{\text{opp } }{ hypot} \\ sin(\bending C) = \frac{half dozen}{ten} \\ sin(\angle C) = \frac{6}{10} \\ sin(\angle C) = .6 $
Trouble 3
What is the cosine ratio of $$\angle C $$ in $$ \triangle ABC$$ ?
Remember the cosine ratio
$ cos(\angle C) =\frac{\text{adj } }{ hypot} \\ cos(C) = \frac{4}{5} \\ cos(C) = .eight $
Trouble four
What is the tangent ratio of $$\angle A$$ ?
Use sochcahtoa to assistance recollect the ratio.
$ tan(A) = \frac{opposite}{adjacent} \\ = \frac{24}{7} \\= 3.42857142 $
Problem 5
Use sochcahtoa to assistance call back the ratios.
$$ \text{ for } \angle \red R \\ \boxed { Sine } \\ sin(\carmine R )= \frac{opp}{hyp} \\ sin(\red R )= \frac{12}{13} \\ sin(\reddish R )= .923 \\ \boxed{ cosine } \\ cos(\blood-red R)= \frac{adj}{hyp} \\ cos(\blood-red R)= \frac{9}{13} = \\ cos(\reddish R).69 \\ \boxed{ tangent } \\ tan(\ruddy R)= \frac{opp}{adj} \\ tan(\red R) = \frac{12}{9} \\ tan(\red R)= 1.3 $$
Trouble half-dozen
What is $$sin(\angle 10)$$?
$$ sin(\red X ) = \frac{opp}{hyp} \\ sin(\cherry X ) = \frac{24}{25} \\ sin(\blood-red 10 ) = .96 $$
Problem seven
What is $$cos(\bending X)$$?
$$ sin(\carmine 10) = \frac{adj}{hyp} \\ sin(\blood-red X) = \frac{7}{25} \\ sin(\scarlet Ten) = .28 $$
Problem 8
Calculate : $$\text{ a) } sin(\angle H) \\ \text{ b) }cos(\angle H) \\ \text{ c) } tan(\angle H)$$
$$ sin(\bending \red H) = \frac{3}{v} \\ sin( \angle \red H) = .6 \\ cos(\angle \red H) =\frac{ 4}{v} \\ cos (\angle \red H)= .8 \\ tan(\angle \blood-red H) = \frac{3}{4} \\ tan(\angle \red H)= .75 $$
elevation | Practice I | #Practice II | Word Problems | Challenge Problem
Practice Bug II
Problem 9
Which angle below has a tangent ratio of $$\frac{three}{iv}$$?
Employ sochcahtoa to help remember the ratio for tangent.
$$ tangent = \frac{opp}{adj} $$
So which angle has a tangent that is equivalent to $$\frac{3}{4}$$?
Yous but accept two options. Either $$\angle Fifty$$ or $$ \bending K $$. So let'southward calculate the tangent for each bending.
$ tan( \bending K) = \frac{12}{ix} \\ \frac{12}{9} \carmine {\ne} \frac three 4 $
$ tan( \angle L) = \frac{nine}{12} \\ \frac{nine}{12} = \frac 3 4 $
Problem 10
Which bending below has a cosine of $$\frac{3}{v}$$?
Use sochcahtoa to aid remember the ratio for tangent.
$$ cosine = \frac{adj}{hyp} $$
Then which angle has a cosine that is equivalent to $$\frac 3 v $$?
You lot only have 2 options. Either $$\angle L$$ or $$ \angle K $$. So allow'southward calculate the tangent for each bending.
$ cos( \angle Thou) = \frac{9}{15} \\ \frac{9}{fifteen} \ruddy {\ne} \frac 3 4 $
$ cos(50) = \frac{12}{fifteen} \\ \frac{12}{15} = \frac 3 5 $
Trouble 11
Which bending below has a tangent $$\approx$$ .29167?
This is a little trickier because you are given the ratio as a decimal; yet, yous only have two options. Either $$ \angle A $$ or $$\angle C $$.
$$ tan(\angle A) = \frac{48}{14} \\ tan(\angle A)= 3.42857 $$
$$ tan(\angle C) = \frac{xiv}{48} \\ tan(\angle C) = .29167 $$
Problem 12
Which angle below has a tangent of 2.iv?
Again, there are two options (angles R or P), but since your ratio is greater than 1 you might speedily be able to notice that it must exist R.
$$ tan(\bending P) = \frac{5}{12} \\ tan(\bending P) = 0. 41666 $$
$$ tan(\angle R) = \frac{12}{5} \\ tan(\bending R) = two.4 $$
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Give-and-take Problems
Problem 13
Step 1
Information technology's easiest to do a word trouble similar this 1, by first drawing the triangle and labelling the sides. We know the opposite side of $$ \angle K$$ and we know the hypotenuse.
To become the tangent ratio we need to know the length of the adjacent side.
How tin we detect the length of the adjacent side?
Stride 2
Use the Pythagorean theorem!
$$ a^2 + b^2 = c^2 \\ iii^2 + b^2 = 5^2 \\ b^two = 5^two- three^2 = 25-nine = 16 \\ b = four $$
At present, use the tangent ratio!
Trouble 14
Step 1
Draw this triangle and label the sides:
Remember that the cosine ratio = $$\frac{adjacent}{hypotenuse}$$.
How can we detect the length of the opposite side?
(Think that sine involves the opposite so we need to find that somehow).
Footstep 2
Use the Pythagorean theorem!
$$ a^2 + b^two = c^2 \\ 7^2 + b^2 = 25^two \\ b^ii = 25^2- 7^ii = 625 - 49 = 576 \\ b = \sqrt{576} =24 $$
Now, apply the sine ratio!
Footstep three
$$ sin(b) = \frac{opposite}{hypotenuse} $$
top | Exercise I | Practice Ii | Word Problems | #Challenge Problem
Challenge Trouble
Exist careful!
In the triangle beneath, which bending has a sine ratio of 2.half dozen?
There is no angle that has a sine of ii.half dozen.
Remember that the maximum value of the sine ratio is ane.
The aforementioned, is truthful, by the way of cosine ratio (max value is ane).
Source: https://www.mathwarehouse.com/trigonometry/sine-cosine-tangent-practice2.php
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