b Use the fact that the hypotenuse is twice the length of the shorter leg to solve for s in terms of h.
C
c Use the fact that the length of the longer leg is sqrt(3) times the length of the shorter leg to solve for s in terms of l.
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a
B
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Practice makes perfect
a In a 30 ° -60 ° -90 ° triangle, the length of the hypotenuse, h, is twice the length of the shorter leg, s. The length of the longer leg, l, is sqrt(3) times the length of the shorter leg.
h= 2s
l= sqrt(3)s
Let's name this △ ABC, where m∠ A=90, m∠ B=60, and m∠ C=30. Let's look at a rough sketch of what our triangle will look like.
To construct △ ABC, we will make the assumption that s= 1 centimeter. This means we will construct AB to be 1 centimeter. Based off the relationship between the side lengths, let's determine what the length of the other sides would be.
BC= 2s=2( 1)=2 cm
CA= sqrt(3)s=sqrt(3)( 1) ≈ 1.73 cm
Start by drawing CA. Use a ruler to ensure it is 1.73 centimeters.
Next, we will draw AB. Use the ruler to measure out 1 centimeter and then as a straight edge to ensure AB is a straight line.
Use a protractor to measure m∠ C=30. Then, use a straight edge to draw the side connecting C and B.
We have successfully constructed our triangle.
b Now let's solve for the shorter leg and longer leg in terms of the hypotenuse, h. Let's start with the shorter leg.
h= 2s ⇒ s= h/2Now let's find the longer leg in terms of the hypotenuse. We know that l= sqrt(3)s. Let's substitute s= h/2 into this equation to find l in terms of h.
If we construct this triangle using the same steps used in Part A, we get the following triangle.
c Finally, let's solve for the shorter leg and the hypotenuse in terms of the longer leg, l. Let's start with the shorter leg. We know from Part A that l= sqrt(3)s. Let's rearrange this to solve for s.
