McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
4. Trigonometry
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Exercise 20 Page 655

Start by identifying the hypotenuse of the right triangle. Then find the sides that are opposite and adjacent to each angle.

Ratios for ∠ J: sin J=2sqrt(5)/5≈ 0.89, cos J=sqrt(5)/5≈ 0.45, tan J=4sqrt(3)/2sqrt(3)=2.00

Ratios for ∠ L: sin L=sqrt(5)/5≈ 0.45, cos L=2sqrt(5)/5≈ 0.89, tan L=2sqrt(3)/4sqrt(3)=0.50

Practice makes perfect

For the given right triangle, we want to write the ratios for the sine, cosine, and tangent of ∠ J and ∠ L.

Ratios for ∠ J

Let's start by identifying the hypotenuse of the triangle and the sides that are opposite and adjacent to ∠ J.

We see that the length of the hypotenuse is 2sqrt(15). The length of the side adjacent to ∠ J is 2sqrt(3) and the length of the side opposite to ∠ J is 4sqrt(3). With this information, we can find the desired ratios.

Ratio Definition Value
sin J Length of leg opposite to∠ J/Length of hypotenuse 4sqrt(3)/2sqrt(15)=2sqrt(5)/5≈ 0.89
cos J Length of leg adjacent to∠ J/Length of hypotenuse 2sqrt(3)/2sqrt(15)=sqrt(5)/5≈ 0.45
tan J Length of leg opposite to∠ J/Length of leg adjacent to∠ J 4sqrt(3)/2sqrt(3)=2.00

Ratios for ∠ L

We already know the length of the hypotenuse is 2sqrt(15). Let's identify the sides that are opposite and adjacent to ∠ L.

The length of the side adjacent to ∠ L is 4sqrt(3) and the length of the side opposite to ∠ L is 2sqrt(3). With this information, we can find the desired ratios.

Ratio Definition Value
sin L Length of leg opposite to∠ L/Length of hypotenuse 2sqrt(3)/2sqrt(15)=sqrt(5)/5≈ 0.45
cos L Length of leg adjacent to∠ L/Length of hypotenuse 4sqrt(3)/2sqrt(15)=2sqrt(5)/5≈ 0.89
tan L Length of leg opposite to∠ L/Length of leg adjacent to∠ L 2sqrt(3)/4sqrt(3)= 0.50

Showing Our Work

Calculations
In case you want to see the steps we took to rationalize the denominators of our ratios, we have included these calculations. We began with 4sqrt(3)2sqrt(15).
4sqrt(3)/2sqrt(15)
4sqrt(3)*sqrt(15)/2sqrt(15)*sqrt(15)
4sqrt(3)*sqrt(3* 5)/2sqrt(15)*sqrt(15)
4sqrt(3)*sqrt(3)*sqrt(5)/2sqrt(15)*sqrt(15)
4* 3*sqrt(5)/2* 15
2* 3*sqrt(5)/15
2sqrt(5)/5
Next we simplified 2sqrt(3)2sqrt(15).
2sqrt(3)/2sqrt(15)
Simplify
sqrt(3)/sqrt(15)
sqrt(3/15)
sqrt(1/5)
sqrt(1)/sqrt(5)
1/sqrt(5)
sqrt(5)/sqrt(5)*sqrt(5)
sqrt(5)/5