Example 1
In this triangle we know the three sides:
Use The Law of Cosines first to find one of the angles. It doesn't matter which one. Let's find angle A first:
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. To do so, we could use The Law of Cosines again, or we could use The Law of Sines. The Law of Sines is a little easier, but we may have made a mistake in our calculation of angle A: so let's use The Law of Cosines to find 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.
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