McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
3. Inequalities in One Triangle
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Exercise 20 Page 431

Start with calculating the measure of ∠ B.

Ben, see solution.

Practice makes perfect

We can begin with copying the triangle from the diagram. Let's name the vertices by the first letters of the players' names.

We are asked to find the shortest passing distance from point H to either G or B. To do this, let's use Theorem 5.10. If one angle of a triangle has a greater measure than another angle, then the side opposite to the greater angle is longer than the side opposite to the lesser angle. From the above diagram, we can see that HG is opposite to ∠ B and HB is opposite to ∠ G. By determining which of these angles is greater, we can find out which side is longer and which is shorter. First, we need to calculate the measure of ∠ B. Let's use the fact that the sum of a triangle's angles is 180^(∘). m∠ B+ m∠ G + m∠ H=180^(∘) We are given that ∠ G measures 48^(∘) and ∠ H measures 62^(∘). Let's substitute these values into the equality and calculate the measure of ∠ B.
m∠ B+ m∠ G + m∠ H=180^(∘)
m∠ B+ 48^(∘) + 62^(∘)=180^(∘)
m∠ B+ 110^(∘)=180^(∘)
m∠ B=70^(∘)
Now we can compare the measures of ∠ G and ∠ B. m∠ G =48^(∘) m∠ B =70^(∘) As we can see, ∠ B has a greater measure. By the theorem, side GH, which is opposite to this angle, is longer. We can conclude that Hannah should pass the ball to Ben, as BH is the shortest passing distance.