McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. Inequalities in One Triangle
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Exercise 47 Page 432

Order the sides of the triangles from shortest to longest. Use Theorem 5.9 for each triangle.

∠ 1, ∠ 2, ∠ 5, ∠ 4, ∠ 6, ∠ 3

Practice makes perfect

We are given that the measure of ∠ 2 is the same as the measure of ∠ 5. Let's add this piece of information to the diagram. We will also name the vertices for the following explanation to be easier.

Now we can recall what Theorem 5.9 states. If one side of a triangle is greater than the other side, then the angle opposite to the longer side has a greater measure than the angle opposite to the shorter side. According to the theorem, instead of comparing the measures of the angles, we can compare the known measures of the sides. Let's start with △ BCD. We can order the sides from shortest to longest as follows. DC, BC, BD Now we need to find the angles opposite to these sides.

DC - ∠ 5 BC - ∠ 4 BD - ∠ 6 Hence, by the above theorem, the order of the angles from smallest to largest is the same as the order of the sides opposite to them. ∠ 5, ∠ 4, ∠ 6 Let's now use the fact that the measure of ∠ 2 is the same as the measure of ∠ 5. This means that we can write ∠ 2 next to ∠ 5. ∠ 2, ∠ 5, ∠ 4, ∠ 6 We can now consider the triangle △ ABD. Similarly, we start with comparing the measures of the sides and writing them in order from the shortest to longest. BD, AD, AB Next, we need to find the angles opposite to these sides. BD - ∠ 1 AD - ∠ 2 AB - ∠ 3 Hence, the order of the angles from smallest to largest is the following. ∠ 1, ∠ 2, ∠ 3 Earlier we established that ∠ 2 and ∠ 5 are less than ∠ 4 and ∠ 6. Since ∠ 1 is even smaller than ∠ 2, it is smaller than all of these angles. Thus, we can write it on the left of the sequence. ∠ 1, ∠ 2, ∠ 5, ∠ 4, ∠ 6 What about ∠ 3? It is larger than ∠ 2 and ∠ 5, but so are ∠ 4 and ∠ 6. To find where to place it in the sequence, let's consider the given diagram once again.

The sum of the measures of a triangle angles is 180^(∘). Therefore, the sum of m∠ 1, m∠ 2, and m∠ 3 is equal to 180^(∘) and the same as the sum of m∠ 4, m∠ 5, and m∠ 6. This allows us to from an equation. m∠ 1+ m∠ 2 + m∠ 3 = m∠ 4+ m∠ 5+ m∠ 6 The angles ∠ 2 and ∠ 5 have the same measure, so we can cross them out the equation and it will remain true. m∠ 1+ m∠ 3 = m∠ 4+ m∠ 6 We have also determined that ∠ 1 is smaller than ∠ 4. Therefore, for the equation to be correct, ∠ 3 must be larger than ∠ 6. We conclude that ∠ 3 should be located to the furthest right of the sequence. ∠ 1, ∠ 2, ∠ 5, ∠ 4, ∠ 6, ∠ 3 The order of the angles from smallest to largest is as written above.