In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Law of Sines.

He will start by defining the Law of Sines and the run through a couple of instances in which we can use the law of sines to solve for unknown characteristics of a triangle.

This Fundamentals of Engineering Exam Review of Law of Sines is part of the global subject Mathematics.

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Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Law of Sines.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Law of Sines by defining exactly what Law of Sines is.

The law of cosines and the law of sines

allows us to solve triangles that are

not right-angled in our called oblique

triangles it states the following the

sides of a triangle are to one another

in the same ratio as the signs of their

opposite angles so to illustrate that

let’s look at the triangle here let’s

say this is side a this is side B side C

this is angle a this is angle C and this

is angle B so the algebraic statement of

the law of sines states that sine of a

over a is equal to sine a B over B which

is equal to the sine of C over C so

let’s take into take and look at example

here let’s look at the triangle let’s

say that we know this angle right here

65 degrees we got this angle 75 degrees

and this angle is 40 degrees let’s say

this side is 10 and let’s say we do not

know this side which we’ll call X and we

want to find that side so all we need to

do is employ the law of sines so again

its sine of a over a is equal to sine of

B over B which is equal to sine of C

over C so in this case we can we can

just take the angle for T in the ankle

75 and the side of 10 and the side 10 to

determine X so let’s go ahead and take

that ratio let’s say the sine of 75 over

10 is going to be equal to the sine of

40

over X now just rearranging and solving

that we get 10 sine of 40 over the sine

of 75 is equal to X and using our

calculators we find that X is equal to

six point six five so what if we are

given a triangle that has an obtuse

angle how do how do we go about solving

that well no differently we do the same

thing we just take the ratios of the

sides and the angles and a soft form so

let’s look at an example there example

let’s say that we have in triangle that

looks something like this and let’s say

this angle is our obtuse angle and it’s

105 and we got this angle which is 25 in

this angle let’s say 50 and let’s say

that we have this side is 10 and this is

our unknown side X so all we need to do

once again is look at our definition of

the law of sines and know that all we

need to do is have plug in a couple few

couple knowns to get that unknown X so

let’s say we got sine of 25 which is the

opposite of our known sine of 10 divided

by 10 is equal to sine of 105 which is

opposite our unknown X and once again

just rearranging that we got X is equal

to 10 sine of 105 divided by sine of 25

solving that using our calculators we

find that X is equal to 22 point eight

so no different just solve it in the

same way just use the ratios so what if

we want to find an unknown angle and all

these all these previous examples we’ve

we kind of just used the law of sines to

determine unknown sides but we can

so you use the law of sines to find the

unknown angle so let’s look at another

example here example let’s say we have a

triangle and let’s say we know this side

is five point five we know this side is

four point seven and we know this side

or this angle is 63 degrees so what we

want to do is find angle B so how do we

do that let’s say this is angle B since

we know side B and we know we know a

side and and that 63 is the angle

opposite to our our second known side

all we need to do again is take our sine

of 63 divided by the known side of

five-and-a-half and our unknowns angle B

divided by our known side of four point

seven so again all we need to do is

rearrange sine of B is equal to four

point seven sine of 63 divided by five

point five and we can take the inverse

sine of this all and let’s calculate

this out this is all getting equal point

seven six one taking that inverse sine

we find that B is equal to forty nine

point six degrees so we’re also able to

find an unknown angle using the law of

sines so that’s it for now guys I hope

you appreciated that quick review that

simple review I know it was probably

pretty basic to you guys but definitely

something that you need to brush off up

on because as easy as it may seem you

will run across some of these questions

on the exam and being able to crank

these out as quick as possible is

definitely going to be of benefit to you

so anyways if you guys have any more

questions about the engineer in training

exam go ahead and visit my site and

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I’m definitely have plans to expand this

channel and get more content up for you

guys to check out so if you have also

any topics that you would like me to

cover in future tutorials

definitely don’t hesitate to contact me

I’m definitely open to all emails and

contacts and I’m just out there to help

you guys out so I’ll do the best that I

can for you so for now take care and I

look forward to talking to you guys soon

all right bye

—

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