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Solving for two unknown sides and angle of an aas triangle

Solve the triangle shown in [link] to the nearest tenth.

An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.

The three angles must add up to 180 degrees. From this, we can determine that

β = 180° 50° 30° = 100°

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c .

sin ( 50° ) 10 = sin ( 30° ) c c sin ( 50° ) 10 = sin ( 30° ) Multiply both sides by  c . c = sin ( 30 ° ) 10 sin ( 50° ) Multiply by the reciprocal to isolate  c . c 6.5

Similarly, to solve for b , we set up another proportion.

    sin ( 50° ) 10 = sin ( 100° ) b    b sin ( 50° ) = 10 sin ( 100° ) Multiply both sides by  b .                 b = 10 sin ( 100° ) sin ( 50° ) Multiply by the reciprocal to isolate  b .                 b 12.9

Therefore, the complete set of angles and sides is

α = 50° a = 10 β = 100° b 12.9 γ = 30° c 6.5
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Solve the triangle shown in [link] to the nearest tenth.

An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.

α = 98 a = 34.6 β = 39 b = 22 γ = 43 c = 23.8

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Using the law of sines to solve ssa triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case    . Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

Possible outcomes for ssa triangles

Oblique triangles in the category SSA may have four different outcomes. [link] illustrates the solutions with the known sides a and b and known angle α .

Four attempted oblique triangles are in a row, all with standard labels. Side c is the horizontal base. In the first attempted triangle, side a is less than the altitude height. Since side a cannot reach side c,  there is no triangle. In the second attempted triangle, side a is equal to the length of the altitude height, so side a forms a right angle with side c. In the third attempted triangle, side a is greater than the altitude height and less than side b, so side a can form either an acute or obtuse angle with side c. In the fourth attempted triangle, side a is greater than or equal to side b, so side a forms an acute angle with side c.

Solving an oblique ssa triangle

Solve the triangle in [link] for the missing side and find the missing angle measures to the nearest tenth.

An oblique triangle with standard labels where side a is of length 6, side b is of length 8, and angle alpha is 35 degrees.

Use the Law of Sines to find angle β and angle γ , and then side c . Solving for β , we have the proportion

sin α a = sin β b sin ( 35° ) 6 = sin β 8 8 sin ( 35° ) 6 = sin β 0.7648 sin β sin 1 ( 0.7648 ) 49.9° β 49.9°

However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β ? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have [link] . It appears that there may be a second triangle that will fit the given criteria.

An oblique triangle built from the previous with standard prime labels. Side a is of length 6, side b is of length 8, and angle alpha prime is 35 degrees. An isosceles triangle is attached, using side a as one of its congruent legs and the angle supplementary to angle beta as one of its congruent base angles. The other congruent angle is called beta prime, and the entire new horizontal base, which extends from the original side c, is called c prime. There is a dotted altitude line from angle gamma prime to side c prime.

The angle supplementary to β is approximately equal to 49.9°, which means that β = 180° 49.9° = 130.1° . (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ , we have

γ = 180° 35° 130.1° 14.9°

We can then use these measurements to solve the other triangle. Since γ is supplementary to γ , we have

γ = 180° 35° 49.9° 95.1°

Now we need to find c and c .

We have

c sin ( 14.9° ) = 6 sin ( 35° )                c = 6 sin ( 14.9° ) sin ( 35° ) 2.7

Finally,

c sin ( 95.1° ) = 6 sin ( 35° )               c = 6 sin ( 95.1° ) sin ( 35° ) 10.4

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in [link] .

There are two triangles with standard labels. Triangle a is the orginal triangle. It has angles alpha of 35 degrees, beta of 130.1 degrees, and gamma of 14.9 degrees. It has sides a = 6, b = 8, and c is approximately 2.7. Triangle b is the extended triangle. It has angles alpha prime = 35 degrees, angle beta prime = 49.9 degrees, and angle gamma prime = 95.1 degrees. It has side a prime = 6, side b prime = 8, and side c prime is approximately 10.4.

However, we were looking for the values for the triangle with an obtuse angle β . We can see them in the first triangle (a) in [link] .

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Questions & Answers

can you not take the square root of a negative number
Sharon Reply
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
Elaine Reply
can I get some pretty basic questions
Ama Reply
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
Amara Reply
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
Mars Reply
what is the domain of f(x)=x-4/x^2-2x-15 then
Conney Reply
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
jeric Reply
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
jeric Reply
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
Abena Reply
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
Ashley Reply
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Fiston Reply
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
Ahlicia Reply
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
Carlos Reply
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
Mary Reply
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
Karim Reply
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
Practice Key Terms 4

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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