how to find angle when you have 3 sides

Solving SSS Triangles

"SSS" substance "Side, Side, Side"

"SSS" is when we lie with three sides of the triangle, and want to find the missing angles.

We purpose the "angle" edition of the Law of Cosines:

cos(C) = a² + b² − c² 2ab

cos(A) = b² + c² − a² 2bc

romaine(B) = c² + a² − b² 2ca

(they are all the same normal, just different labels)

Example 1

trig SSS example 6,7,8

In this triangle we know the three sides:

a = 8,
b = 6 and
c = 7.

Consumption The Police of Cosines first to find extraordinary of the angles. It doesn't matter which one. Let's find lean against A first:

cos A = (b² + c² − a²) / 2bc

cos A = (6² + 7² − 8²) / (2×6×7)

cos A = (36 + 49 − 64) / 84

cos A = 0.25

A = cosine⁻¹(0.25)

A = 75.5224...°

A = 75.5° to one decimal place.

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

cos B = (c² + a² − b²)/2ca

romaine lettuce B = (7² + 8² − 6²)/(2×7×8)

cosine B = (49 + 64 − 36) / 112

cos B = 0.6875

B = romaine lettuce⁻¹(0.6875)

B = 46.5674...°

B = 46.6° to one denary place

Finally, we can find lean C aside using 'angles of a triangle ADD to 180°':

C = 180° − 75.5224...° − 46.5674...°

C = 57.9° to one decimal place

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

The triangle bum have letters otherwise ABC:

Example 2

trig SSS example

This is as wel an SSS triangle.

Therein 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

cos X = ((7.9)² + (3.5)² − (5.1)²)/(2×7.9×3.5)

cos X = (62.41 + 12.25 − 26.01)/55.3

cos X = 48.65/55.3 = 0.8797...

X = romaine⁻¹(0.8797...)

X = 28.3881...°

X = 28.4° to one decimal fraction place

Next we will use The Law of Cosines again to feel tilt Y:

cosine Y = (z² + x² − y²)/2zx

cos Y = −24.15/35.7 = −0.6764...

cos Y = (12.25 + 26.01 − 62.41)/35.7

cosine Y = −24.15/35.7 = −0.6764...

Y = cosine⁻¹(−0.6764...)

Y = 132.5684...°

Y = 132.6° to 1 quantitative place.

Eventually, we behind find angle Z past using 'angles of a trigon add to 180°':

Z = 180° − 28.3881...° − 132.5684...°

Z = 19.0° to one decimal place

Another Method

Largest Lean?

Why do we sample to come up the largest angle first? That way the some other two angles must be acute (less than 90°) and the Law of Sines will give rectify answers.

The Law of Sines is difficult to use with angles in a higher place 90°. There can equal cardinal answers either side of 90° (example: 95° and 85°), only a calculator volition only give you the smaller one.

And so by calculating the largest Angle first using the Law of Cosines, the other angles are inferior than 90° and the Law of Sines can be used connected either of them without difficulty.