Solving SSS Triangles

"SSS" means "Side, Side, Side"

When you know three sides of the triangle, and want to find the missing angles.

 

To solve an SSS triangle: use The Law of Cosines first to calculate one of the angles then use The Law of Cosines again to find another angle and finally use angles of a triangle add to 180° to find the last angle.

 
Example 1



In this triangle we know the three sides:

a = 8,

b = 6 and

c = 7.


 
Use The Law of Cosines first to find one of the angles. It doesn't matter which one. Let's find angle Afirst:

cos A = (b2 + c2 - a2)/2bc

= (62 + 72 - 82)/(2×6×7) = (36 + 49 - 64)/84 = 21/84 = 0.25

A = cos-1(0.25)

= 75.5° correct to one decimal place.

Next we will find another side. We use The Law of Cosines again, this time for angle B:

cos B = (c2 + a2 - b2)/2ca

= (72 + 82 - 62)/(2×7×8) = (49 + 64 - 36)/112 = 77/112 = 0.6875

So B = 46.5674...°

= 46.6° correct to one decimal place.

Finally, we can find angle C by using 'angles of a triangle add to 180°':

So C = 180° - 75.5224...° - 46.5674...°

= 57.9° correct to one decimal place.

Now we have completely solved the triangle i.e. we have found all its angles.


 
Example 2



This is also an SSS triangle.

In this triangle we know the three sides x = 5.1, y = 7.9 and z = 3.5. Use The Law of Cosines to find angle X first:

cos X = (y2 + z2 - x2)/2yz

= ((7.9)2 + (3.5)2 - (5.1)2)/(2×7.9×3.5)

= (62.41 + 12.25 - 26.01)/55.3

= 48.65/55.3 = 0.8797...

So X = cos-1(0.8797...)

= 28.3881...°

= 28.4° correct to one decimal place.

Next we will use The Law of Cosines again to find angle Y:

cosY = (z2 + x2 - y2)/2zx

= ((3.5)2 + (5.1)2 - (7.9)2)/(2×3.5×5.1)

= (12.25 + 26.01 - 62.41)/35.7

= -24.15/35.7 = -0.6764...

So Y = cos-1(-0.6764...)

= 132.5684...°

= 132.6° correct to one decimal place.

Finally, we can find angle Z by using 'angles of a triangle add to 180°':

So Z = 180° - 28.3881...° - 132.5684...°   

= 19.0° correct to one decimal place.

Another Method

Here is another (slightly faster) way to solve an SSS triangle: use The Law of Cosines first to calculate the largest angle then use The Law of Sines to find another angle and finally use angles of a triangle add to 180° to find the last angle.

Largest Angle?

Why do we try to find the largest angle first? That way the other two angles must be acute (less than 90°) and the Law of Sines will give correct answers.

You see, the Law of Sines is difficult to use with angles above 90°. There can be two answers either side of 90° (example: 95° and 85°), but your calculator will only give you the smaller one.

So by calculating the largest angle first using the Law of Cosines, the remaining angles will be less than 90° and the Law of Sines can be used on either of them without difficulty.

Example 3



B is the largest angle, so find B first using the Law of Cosines:

cos B = (a2 + c2 – b2) / 2ac

cos B = (11.62 + 7.42 – 15.22) / (2×11.6×7.4)

cos B = (134.56 + 54.76 – 231.04) / 171.68

cos B = -41.72 / 171.68

cos B = -0.2430...

B = 104.1°

Use the Law of Sines, sinC/c = sinB/b, to find angle A:

sin C / 7.4 = sin 104.1° / 15.2

sin C = (7.4 sin 104.1°) / 15.2 = 0.4722...

C = 28.2°

Find angle A using "angles of a triangle add to 180":

A = 180° - (104.1° + 28.2°)

A = 180° - 132.3°

A = 47.7°


 
Therefore A = 47.7°, B = 104.1°, and C = 28.2°


 

Pythagorean Theorem

One of the most famous mathematicians who has ever lived, Pythagoras, a Greek scholar who lived way back in the 6th century B.C. (back when Bob Dole was learning geometry), came up with one of the most famous theorems ever, the Pythagorean Theorem.  It says - in a right triangle, the square of the measure of the hypotenuse equals the sum of the squares of the measures of the two legs.  This theorem is normally represented by the following equation: a2 + b2 = c2, where c represents the hypotenuse.

With this theorem, if you are given the measures of two sides of a triangle, you can easily find the measure of the other side.



1. Problem: Find the value of c.


 



 
Solution: a2 + b2 = c2 Write the Pythagorean

Theorem and then plug in any

given information.


 
52 + 122 = c2 The information that was

given in the figure was

plugged in.


 

 
169 = c2 Solve for c

c = 13




 


One of the special right triangles which we deal with in geometry is an isosceles right triangle.  These triangles are also known as 45-45-90 triangles (so named because of the measures of their angles).  There is one theorem that applies to these triangles.  It is stated below.

In a 45-45-90 triangle, the measure of the hypotenuse is equal to the measure of a leg multiplied by SQRT(2).

The following figure presents the theorem in graphical terms.






 


There's another kind of special right triangle which we deal with all the time.  These triangles are known as 30-60-90 triangles (so named because of the measures of their angles).  There is one theorem that applies to these triangles.  It is stated below.

In a 30-60-90 triangle, the measure of the hypotenuse is two times that of the leg opposite the 30o angle.  The measure of the other leg is SQRT(3) times that of the leg opposite the 30o angle.

The following figure presents the theorem in graphical terms.






While the word trigonometry strikes fear into the hearts of many, we made it through (amazing as it may seem to us), and hope to help you through it, too!  Each of the three basic trigonometric ratios are shown below.



sine of angle A = (measure of opposite leg)/(measure of hypotenuse).  In the figure, the sin of angle A = (a/c).

cosine of angle A = (measure of adjacent leg)/(measure of hypotenuse).  In the figure, the cos of angle A = (b/c).

tangent of angle A = (measure of opposite leg)/(measure of adjacent leg).  In the figure, the tan of angle A = (a/b).



1. Problem: Find sin A, cos A, and tan A.


  
Solution: sine = (opposite/hypotenuse)   

sine = 5/13


 

 
cosine = (adjacent/hypotenuse)

cos = 12/13


 
tangent = (opposite/adjacent)

tan = 5/12

Be aware that, although the example above seems to indicate otherwise, the values for the trigonometric ratios depend on the measure of the angle, not the measures of the triangle's sides.


Many problems ask that you find the measure of an angle or a segment that cannot easily be measured.  Problems of this kind can often be solved by the application of trigonometry.  Below is an example problem of this type.

1. Problem: A ladder 12 meters long leans

against a building. It rests on

the wall at a point 10 meters

above the ground. Find the angle

the ladder makes with the ground.


 
Solution: Make sure you know what is being

asked. Then use the given

information to draw and label a

figure. Here's our idea of a

figure for this problem:


  
Choose a variable to represent the

measure of the angle you are asked

to find. Using the variable you

have chosen, write an equation that

will solve the problem.


 
sin x2 = (10/12)


 
The above equation is derived from

the given information and the

knowledge of the sine

ratio.


 
Find the solution using a calculator's

Arcsine function or a table

of trigonometric ratios.


 
TI-82 screen: sin-1 (10/12) = 56.44


 
Trigonometric Ratios Table:

sin 56o = 0.8290

sin 57o = 0.8387


 
By either answer, after rounding to

the nearest degree, the answer is 56o.