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McGraw Hill Glencoe Geometry, 2012 View details

4. Trigonometry

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Exercise 24 Page 574

Draw and label the side lengths of a 45^(∘)-45^(∘)-90^(∘) triangle.

sqrt(2)/2 or approximately 0.71

Practice makes perfect

Let's begin with drawing a 45^(∘)-45^(∘)-90^(∘) triangle. If we call the legs of this right isosceles triangle x, then the hypotenuse is xsqrt(2).

In our exercise we are asked to evaluate the sine of 45^(∘). Recall that the sine of ∠ A is the ratio of the length of the leg opposite ∠ A to the length of the hypotenuse. Using this definition, we can write the appropriate ratio for 45^(∘).

sin 45^(∘)=x/xsqrt(2)

ReduceFrac

a/b=.a /x./.b /x.

sin 45^(∘)=1/sqrt(2)

▼

Simplify right-hand side

ExpandFrac

a/b=a * sqrt(2)/b * sqrt(2)

sin 45^(∘)=1*sqrt(2)/sqrt(2)*sqrt(2)

ProdToPowTwoFac

a* a=a^2

sin 45^(∘)=1*sqrt(2)/(sqrt(2))^2