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Geometry (all content)

Course: Geometry (all content) > Unit 13

Lesson 11: Solving general triangles
Math
Geometry (all content)
Trigonometry
Solving general triangles
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Laws of sines and cosines review

Google Classroom
Review the law of sines and the law of cosines, and use them to solve problems with any triangle.

Law of sines

start fraction, a, divided by, sine, left parenthesis, alpha, right parenthesis, end fraction, equals, start fraction, b, divided by, sine, left parenthesis, beta, right parenthesis, end fraction, equals, start fraction, c, divided by, sine, left parenthesis, gamma, right parenthesis, end fraction

Law of cosines

c, squared, equals, a, squared, plus, b, squared, minus, 2, a, b, cosine, left parenthesis, gamma, right parenthesis
Want to learn more about the law of sines? Check out this video.
Want to learn more about the law of cosines? Check out this video.

Practice set 1: Solving triangles using the law of sines

This law is useful for finding a missing angle when given an angle and two sides, or for finding a missing side when given two angles and one side.

Example 1: Finding a missing side

Let's find A, C in the following triangle:
According to the law of sines, start fraction, A, B, divided by, sine, left parenthesis, angle, C, right parenthesis, end fraction, equals, start fraction, A, C, divided by, sine, left parenthesis, angle, B, right parenthesis, end fraction. Now we can plug the values and solve:
ABsin⁡(∠C)=ACsin⁡(∠B)5sin⁡(33∘)=ACsin⁡(67∘)5sin⁡(67∘)sin⁡(33∘)=AC8.45≈AC\begin{aligned} \dfrac{AB}{\sin(\angle C)}&=\dfrac{AC}{\sin(\angle B)} \\\\ \dfrac{5}{\sin(33^\circ)}&=\dfrac{AC}{\sin(67^\circ)}\\\\ \dfrac{5\sin(67^\circ)}{\sin(33^\circ)}&=AC \\\\ 8.45&\approx AC \end{aligned}

Example 2: Finding a missing angle

Let's find m, angle, A in the following triangle:
According to the law of sines, start fraction, B, C, divided by, sine, left parenthesis, angle, A, right parenthesis, end fraction, equals, start fraction, A, B, divided by, sine, left parenthesis, angle, C, right parenthesis, end fraction. Now we can plug the values and solve:
BCsin⁡(∠A)=ABsin⁡(∠C)11sin⁡(∠A)=5sin⁡(25∘)11sin⁡(25∘)=5sin⁡(∠A)11sin⁡(25∘)5=sin⁡(∠A)\begin{aligned} \dfrac{BC}{\sin(\angle A)}&=\dfrac{AB}{\sin(\angle C)} \\\\ \dfrac{11}{\sin(\angle A)}&=\dfrac{5}{\sin(25^\circ)} \\\\ 11\sin(25^\circ)&=5\sin(\angle A) \\\\ \dfrac{11\sin(25^\circ)}{5}&=\sin(\angle A) \end{aligned}
Evaluating using the calculator and rounding:
m, angle, A, equals, sine, start superscript, minus, 1, end superscript, left parenthesis, start fraction, 11, sine, left parenthesis, 25, degrees, right parenthesis, divided by, 5, end fraction, right parenthesis, approximately equals, 68, point, 4, degrees
Remember that if the missing angle is obtuse, we need to take 180, degrees and subtract what we got from the calculator.
Problem 1.1
  • Current
B, C, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Round to the nearest tenth.

Want to try more problems like this? Check out this exercise.

Practice set 2: Solving triangles using the law of cosines

This law is mostly useful for finding an angle measure when given all side lengths. It's also useful for finding a missing side when given the other sides and one angle measure.

Example 1: Finding an angle

Let's find m, angle, B in the following triangle:
According to the law of cosines:
left parenthesis, A, C, right parenthesis, squared, equals, left parenthesis, A, B, right parenthesis, squared, plus, left parenthesis, B, C, right parenthesis, squared, minus, 2, left parenthesis, A, B, right parenthesis, left parenthesis, B, C, right parenthesis, cosine, left parenthesis, angle, B, right parenthesis
Now we can plug the values and solve:
(5)2=(10)2+(6)2−2(10)(6)cos⁡(∠B)25=100+36−120cos⁡(∠B)120cos⁡(∠B)=111cos⁡(∠B)=111120\begin{aligned} (5)^2&=(10)^2+(6)^2-2(10)(6)\cos(\angle B) \\\\ 25&=100+36-120\cos(\angle B) \\\\ 120\cos(\angle B)&=111 \\\\ \cos(\angle B)&=\dfrac{111}{120} \end{aligned}
Evaluating using the calculator and rounding:
m, angle, B, equals, cosine, start superscript, minus, 1, end superscript, left parenthesis, start fraction, 111, divided by, 120, end fraction, right parenthesis, approximately equals, 22, point, 33, degrees

Example 2: Finding a missing side

Let's find A, B in the following triangle:
According to the law of cosines:
left parenthesis, A, B, right parenthesis, squared, equals, left parenthesis, A, C, right parenthesis, squared, plus, left parenthesis, B, C, right parenthesis, squared, minus, 2, left parenthesis, A, C, right parenthesis, left parenthesis, B, C, right parenthesis, cosine, left parenthesis, angle, C, right parenthesis
Now we can plug the values and solve:
(AB)2=(5)2+(16)2−2(5)(16)cos⁡(61∘)(AB)2=25+256−160cos⁡(61∘)AB=281−160cos⁡(61∘)AB≈14.3\begin{aligned} (AB)^2&=(5)^2+(16)^2-2(5)(16)\cos(61^\circ) \\\\ (AB)^2&=25+256-160\cos(61^\circ) \\\\ AB&=\sqrt{281-160\cos(61^\circ)} \\\\ AB&\approx 14.3 \end{aligned}
Problem 2.1
  • Current
m, angle, A, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
degrees
Round to the nearest degree.

Want to try more problems like this? Check out this exercise.

Practice set 3: General triangle word problems

Problem 3.1
  • Current
"Only one remains." Ryan signals to his brother from his hiding place.
Matt nods in acknowledgement, spotting the last evil robot.
"34 degrees." Matt signals back, informing Ryan of the angle he observed between Ryan and the robot.
Ryan records this value on his diagram (shown below) and performs a calculation. Calibrating his laser cannon to the correct distance, he stands, aims, and fires.
To what distance did Ryan calibrate his laser cannon?
Do not round during your calculations. Round your final answer to the nearest meter.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
start text, space, m, end text

Want to try more problems like this? Check out this exercise.

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  • blobby green style avatar for user victoria
    Posted 3 years ago. Direct link to victoria's post “I want to know why this a...”
    I want to know why this article says "Remember that if the missing angle is obtuse, we need to take 180 degrees and subtract what we got from the calculator" when using the law of sines to find a missing angle. Are there videos that explain why this is? I don't understand why during calculations it won't just give the obtuse angle, and instead gives an acute angle.

    I'm only taking this geometry course and there is nothing along the way that explains this, not even the videos that introduce the law of sines.

    Thanks.
    (19 votes)
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    • leaf green style avatar for user cossine
      Posted 3 years ago. Direct link to cossine's post “There can be two since si...”
      Good Answer
      There can be two since sin(theta) = sin(180-theta) for all values of theta that are real numbers e.g. -1000.98, sqrt(2) etc.

      Since you are using the sin^-1 function you will only ever get 1 angle as the range is defined from -90 to 90 degrees(which is -pi/2 to pi/2 in radians). You can it sketch on desmos to see what it look likes.

      So if you need to find an obtuse angle then you need to use 180-theta to find the obtuse angle.

      It might be helpful to take look at videos related function as well.
      (18 votes)
  • cacteye green style avatar for user Micron
    Posted 6 years ago. Direct link to Micron's post “in problem 3.3 I had to o...”
    in problem 3.3 I had to open the explanation and do not understand why the law of sines in this solution is switched to sin(c)/length of side. isnt it usually the other way around??
    (11 votes)
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  • leafers sapling style avatar for user 3.1415926.tp
    Posted 7 years ago. Direct link to 3.1415926.tp's post “I can't find anything her...”
    I can't find anything here about ambiguous triangles. What if a question asks you to solve from a description where two triangles exist? Like "Determine the unknown side and angles in each triangle, if two solutions are possible, give both: In triangle ABC, <C = 31, a = 5.6, and c = 3.9." I solved for height and see that two solutions exist, and the answer key in my textbook agrees, but I can't figure out how to get either. From a set of questions that's only supposed to be on sine law.
    (5 votes)
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    • primosaur ultimate style avatar for user Jordan Cooper
      Posted 7 years ago. Direct link to Jordan Cooper's post “Use the Law of Sines to g...”
      Good Answer
      Use the Law of Sines to get one possible angle A:
      sin(A)/a=sin(C)/c
      sin(A)/5.6=sin(31)/3.9
      sin(A)=5.6sin(31)/3.9
      A=arcsin(5.6sin(31)/3.9)=47.6924

      Subtract 31 (C) and this angle (A) from 180 to find the third angle (B=101.3076) and use the Law of Sines again to find the third side. If you use the given angle-side pair (C and c) you will be less likely to incur error from your own rounding of angle A:
      b/sin(B)=c/sin(C)
      b/sin(101.3076)=3.9/sin(31)
      b=3.9sin(101.3076)/sin(31)=7.4253

      But if you know that supplementary angles share a sine value, you know that A can also be an obtuse angle with the same sine as 47.6924:
      A=180-47.6924=132.3076

      And again, subtract 31 (C) and the obtuse angle A from 180 to find the other possible third angle (B=16.6924) and use the Law of Sines to find the other possible third side, again using angle C and side c to avoid errors from rounding:
      b/sin(B)=c/sin(C)
      b/sin(16.6924)=3.9/sin(31)
      b=3.9sin(16.6924)/sin(31)=2.1750

      It all comes from knowing that there are two angles, one obtuse and one acute, for every sine value. And you find the obtuse one by subtracting the acute one from 180.

      If you try to do this with a unique triangle (one without two possible sets of angles and sides) your given angle and the obtuse angle you find will add up to more than 180, and so if you try to find a third angle to go with the obtuse one, your subtraction will tell you the third angle is negative, at which point you know you're going down a nonsensical mathematical road, and there weren't two possible triangles to begin with.

      Good luck!
      (17 votes)
  • duskpin ultimate style avatar for user Jonathan
    Posted 4 years ago. Direct link to Jonathan's post “So, obviously, there is t...”
    So, obviously, there is the law of sines and the law of cosines. That is what this entire section has been about. However, I'm curious about if there is such a thing as the law of tangents. Since there is both sine and cosine, wouldn't it make sense if there was something like the law of tangents?
    (8 votes)
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  • winston baby style avatar for user Utkarsh Sharma
    Posted 3 years ago. Direct link to Utkarsh Sharma's post “In the Video 'Solving An ...”
    In the Video 'Solving An Angle With The Law Of Sines' at
    2:05
    , Sal said that Law of Sines is 'sin(a)/A = sin(b)/B = sin(c)/C (lower case letters are the angles and the upper case letters are the side opposite to the angle), in this article it says 'A/sin(a) = B/sin(b) = C/sin(c)'. So which one should I choose?
    (4 votes)
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    • eggleston blue style avatar for user Jeff Dodds
      Posted 3 years ago. Direct link to Jeff Dodds's post “The Law of Sines can be w...”
      The Law of Sines can be written either way! You can put the angles in the numerators and the sides in the denominators, or the other way around.

      To understand why, think about this true equation:

      1/2 = 2/4 = 3/6

      It's still true if we reverse the numerators and denominators:

      2/1 = 4/2 = 6/3
      (5 votes)
  • aqualine sapling style avatar for user Joshua
    Posted 6 years ago. Direct link to Joshua's post “If law of sines is a/sin(...”
    If law of sines is a/sin(a)=b/sin(b)=c/sin(c), does sin(a)/a=sin(b)/b=sin(c)/c work?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user veliwalabasavanarendra
    Posted 6 years ago. Direct link to veliwalabasavanarendra's post “How to find out length of...”
    How to find out length of the segments of a diagonal of quadrilateral.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • primosaur ultimate style avatar for user Jordan Cooper
      Posted 6 years ago. Direct link to Jordan Cooper's post “Two triangles can be iden...”
      Two triangles can be identified in a quadrilateral with one diagonal drawn. Eight triangles can be identified in a quadrilateral with both diagonals drawn. With the diagonal or diagonals drawn, look for a triangle with enough side and angle measures that you can use the law of sines or law of cosines. Doing so may give you enough information to complete other triangles until you have the measurements you want.
      (3 votes)
  • purple pi teal style avatar for user vv1101
    Posted 3 years ago. Direct link to vv1101's post “can someone explain me ho...”
    can someone explain me how Sal gets from this:36=196+144−336cos(∠A)
    to this: 336cos(∠A)=304
    (3 votes)
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  • blobby blue style avatar for user tal-karmon
    Posted 2 years ago. Direct link to tal-karmon's post “How would I use the law o...”
    How would I use the law of cosines to solve for a second side if I know two angles and a side? I couldn't find a video that covers that.
    (3 votes)
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    • female robot grace style avatar for user loumast17
      Posted 2 years ago. Direct link to loumast17's post “To use ONLY the law of co...”
      To use ONLY the law of cosines you needthree of the four parts. The four parts are three sides and one angle. So you either need 2 sides and an angle to solve for the remaining side or all three sides to solve for an angle.

      So if you know two angles (which lets you figure out the third so technically three angles) and a side you can't just use the law of cosines. You need to use another method to solve for one other side. law of sines would do that easily here.
      (3 votes)
  • hopper cool style avatar for user nandhu27
    Posted 3 years ago. Direct link to nandhu27's post “What is the numeric form ...”
    What is the numeric form of sine, cosine, and tangent? Like what does the calculator use to calculate trig functions?
    (3 votes)
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    • leaf green style avatar for user kubleeka
      Posted 3 years ago. Direct link to kubleeka's post “The trig functions can be...”
      The trig functions can be expressed as polynomials of infinite degree, called Taylor polynomials. For example, sin(x)=x-x³/6 +x⁵/120 -x⁷/5040+... Some calculators use these series to approximate the values of the trig functions, as well as other functions like exponentials.

      You'll develop Taylor polynomials in calculus.
      (2 votes)