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# 5.7: Angle-Angle-Side Triangles

Difficulty Level: At Grade Created by: CK-12
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You and a friend decide to go fly kites on a breezy Saturday afternoon. While sitting down to make your kites, you are working on make the best shape possible to catch the breeze. While your friend decides to go with a diamond shaped kite, you try out making a triangle shaped one. While trying to glue the kite together, you make the first and second piece lock together with a 70\begin{align*}70^\circ\end{align*} angle. The angle between the first and third pieces is 40\begin{align*}40^\circ\end{align*}. Finally, you also have measured the length of the second piece and found that it is 22 inches long.

Is there a way to find out, using math, what the length of the third side will be?

Keep reading, and you'll be able to answer this question at the end of this Concept.

### Watch This

James Sousa Example: Solving a Triangle Using the Law of Sines Given Two Angles and One Side

### Guidance

The Law of Sines states: sinAa=sinBb\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}. This is a ratio between the sine of an angle in a triangle and the length of the side opposite that angle to the sine of a different angle in that triangle and the length of the side opposing that second angle.

The Law of Sines allows us to find many quantities of interest in triangles by comparing sides and interior angles as a ratio. One case where we can to use the Law of Sines is when we know two of the angles in a triangle and a non-included side (AAS).

#### Example A

Using GMN,G=42,N=73\begin{align*}\triangle{GMN}, \angle{G} = 42^\circ, \angle{N} = 73^\circ\end{align*} and g=12\begin{align*}g = 12\end{align*}. Find n\begin{align*}n\end{align*}.

Since we know two angles and one non-included side (g)\begin{align*}(g)\end{align*}, we can find the other non-included side (n)\begin{align*}(n)\end{align*}.

sin73nnsin42nn=sin4212=12sin73=12sin73sin4217.15

#### Example B

Continuing on from Example A, find M\begin{align*}\angle{M}\end{align*} and m\begin{align*}m\end{align*}.

Solution: M\begin{align*}\angle{M}\end{align*} is simply 1804273=65\begin{align*}180^\circ - 42^\circ - 73^\circ = 65^\circ\end{align*}. To find side m\begin{align*}m\end{align*}, you can now use either the Law of Sines or Law of Cosines. Considering that the Law of Sines is a bit simpler and new, let’s use it. It does not matter which side and opposite angle you use in the ratio with M\begin{align*}\angle{M}\end{align*} and m\begin{align*}m\end{align*}.

Option 1: G\begin{align*}\angle{G}\end{align*} and g\begin{align*}g\end{align*}

sin65mmsin42mm=sin4212=12sin65=12sin65sin4216.25

Option 2: N\begin{align*}\angle{N}\end{align*} and n\begin{align*}n\end{align*}

sin65mmsin73mm=sin7317.15=17.15sin65=17.15sin65sin7316.25

#### Example C

A business group wants to build a golf course on a plot of land that was once a farm. The deed to the land is old and information about the land is incomplete. If AB\begin{align*}AB\end{align*} is 5382 feet, BC\begin{align*}BC\end{align*} is 3862 feet, AEB\begin{align*}\angle{AEB}\end{align*} is 101,BDC\begin{align*}101^\circ,\angle{BDC}\end{align*} is 74,EAB\begin{align*}74^\circ,\angle{EAB}\end{align*} is 41\begin{align*}41^\circ\end{align*} and DCB\begin{align*}\angle{DCB}\end{align*} is 32\begin{align*}32^\circ\end{align*}, what are the lengths of the sides of each triangular piece of land? What is the total area of the land?

Solution: Before we can figure out the area of the land, we need to figure out the length of each side. In ABE\begin{align*}\triangle ABE\end{align*}, we know two angles and a non-included side. This is the AAS case. First, we will find the third angle in ABE\begin{align*}\triangle ABE\end{align*} by using the Triangle Sum Theorem. Then, we can use the Law of Sines to find both AE\begin{align*}AE\end{align*} and EB\begin{align*}EB\end{align*}.

ABEsin1015382AE(sin101)AEAE=180(41+101)=38=sin38AE=5382(sin38)=5382(sin38)sin101=3375.5 feetsin1015382=sin41EBEB(sin101)=5382(sin41)EB=5382(sin41)sin101EB3597.0 feet

Next, we need to find the missing side lengths in DCB\begin{align*}\triangle DCB\end{align*}. In this triangle, we again know two angles and a non-included side (AAS), which means we can use the Law of Sines. First, let’s find DBC=180(74+32)=74\begin{align*}\angle{DBC} = 180 - (74 + 32) = 74^\circ\end{align*}. Since both BDC\begin{align*}\angle{BDC}\end{align*} and DBC\begin{align*}\angle{DBC}\end{align*} measure 74\begin{align*}74^\circ\end{align*}, DCB\begin{align*}\triangle DCB\end{align*} is an isosceles triangle. This means that since BC\begin{align*}BC\end{align*} is 3862 feet, DC\begin{align*}DC\end{align*} is also 3862 feet. All we have left to find now is DB\begin{align*}DB\end{align*}.

sin743862DB(sin74)DBDB=sin32DB=3862(sin32)=3862(sin32)sin742129.0 feet

Finally, we need to calculate the area of each triangle and then add the two areas together to get the total area. From the last section, we learned two area formulas, K=12 bcsinA\begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} and Heron’s Formula. In this case, since we have enough information to use either formula, we will use K=12 bcsinA\begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} since it is less computationally intense.

First, we will find the area of ABE\begin{align*}\triangle ABE\end{align*}.

ABE\begin{align*}\triangle ABE\end{align*}:

KK=12(3375.5)(5382)sin41=5,959,292.8 ft2

DBC\begin{align*}\triangle DBC\end{align*}:

KK=12(3862)(3862)sin32=3,951,884.6 ft2

The total area is 5,959,292.8+3,951,884.6=9,911,177.4 ft2\begin{align*}5,959,292.8 + 3,951,884.6 = 9,911,177.4\ ft^2\end{align*}.

### Guided Practice

1. Find side "d" in the triangle below with the following information: e=214.9,D=39.7,E=41.3\begin{align*}e = 214.9, D = 39.7^\circ, E = 41.3^\circ\end{align*}

2. Find side "o" in the triangle below with the following information: M=31,O=9,m=15\begin{align*}M = 31^\circ, O = 9^\circ, m = 15\end{align*}

3. Find side "q" in the triangle below with the following information: Q=127,R=21.8,r=3.62\begin{align*}Q = 127^\circ, R = 21.8^\circ, r = 3.62\end{align*}

Solutions:

1. sin41.3214.9=sin39.7d,d=208.0\begin{align*}\frac{\sin 41.3^\circ}{214.9} = \frac{\sin 39.7^\circ}{d}, d = 208.0\end{align*}

2. sin9o=sin3115,o=4.6\begin{align*}\frac{\sin 9^\circ}{o} = \frac{\sin 31^\circ}{15}, o = 4.6\end{align*}

3. sin127q=sin21.83.62,q=7.8\begin{align*}\frac{\sin 127^\circ}{q} = \frac{\sin 21.8^\circ}{3.62}, q = 7.8\end{align*}

### Concept Problem Solution

Since you know two angles and one non-included side of the kite, you can find the other non-included side using the Law of Sines. Set up a ratio using the angles and side you know and the side you don't know.

sin70xxx=sin4022=22sin70sin4032.146

The length of the dowel rod on the unknown side will be approximately 32 inches.

### Explore More

In ABC\begin{align*}\triangle ABC\end{align*}, mA=50\begin{align*}m\angle A=50^\circ\end{align*}, mB=34\begin{align*}m\angle B=34^\circ\end{align*}, and a=6.

1. Find the length of b.
2. Find the length of c.

In KMS\begin{align*}\triangle KMS\end{align*}, mK=42\begin{align*}m\angle K=42^\circ\end{align*}, mM=26\begin{align*}m\angle M=26^\circ\end{align*}, and k=14.

1. Find the length of m.
2. Find the length of s.

In DEF\begin{align*}\triangle DEF\end{align*}, mD=52\begin{align*}m\angle D=52^\circ\end{align*}, mE=78\begin{align*}m\angle E=78^\circ\end{align*}, and d=23.

1. Find the length of e.
2. Find the length of f.

In PQR\begin{align*}\triangle PQR\end{align*}, mP=2\begin{align*}m\angle P=2^\circ\end{align*}, mQ=79\begin{align*}m\angle Q=79^\circ\end{align*}, and p=20.

1. Find the length of q.
2. Find the length of r.

In DOG\begin{align*}\triangle DOG\end{align*}, mD=50\begin{align*}m\angle D=50^\circ\end{align*}, mG=59\begin{align*}m\angle G=59^\circ\end{align*}, and o=12.

1. Find the length of d.
2. Find the length of g.

In CAT\begin{align*}\triangle CAT\end{align*}, mC=82\begin{align*}m\angle C=82^\circ\end{align*}, mT=4\begin{align*}m\angle T=4^\circ\end{align*}, and a=8.

1. Find the length of c.
2. Find the length of t.

In YOS\begin{align*}\triangle YOS\end{align*}, mY=65\begin{align*}m\angle Y=65^\circ\end{align*}, mO=72\begin{align*}m\angle O=72^\circ\end{align*}, and s=15.

1. Find the length of o.
2. Find the length of y.

In HCO\begin{align*}\triangle HCO\end{align*}, mH=87\begin{align*}m\angle H=87^\circ\end{align*}, mC=14\begin{align*}m\angle C=14^\circ\end{align*}, and o=19.

1. Find the length of h.
2. Find the length of c.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.7.

### Vocabulary Language: English

Angle Angle Side Triangle

Angle Angle Side Triangle

An 'angle angle side triangle' is a triangle where two of the angles and the non-included side are known quantities.

## Date Created:

Sep 26, 2012

Feb 26, 2015
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