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EXAMPLE 1 Using a 45 o –45 o –90 o Triangle Softball The infield of a softball field is a square with a side length of 60 feet. A catcher throws the ball.

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Presentation on theme: "EXAMPLE 1 Using a 45 o –45 o –90 o Triangle Softball The infield of a softball field is a square with a side length of 60 feet. A catcher throws the ball."— Presentation transcript:

1 EXAMPLE 1 Using a 45 o –45 o –90 o Triangle Softball The infield of a softball field is a square with a side length of 60 feet. A catcher throws the ball from home plate to second base. How far does the catcher have to throw the ball? SOLUTION To find the distance, use the rule for a 45 o –45 o –90 o triangle.

2 EXAMPLE 1 hypotenuse = leg 2 = 60 2 60(1.414) = 84.84 A catcher has to throw the ball about 85 feet. ANSWER Using a 45 o –45 o –90 o Triangle

3 GUIDED PRACTICE for Example 1 1. A city park is shaped like a square with a side length of 150 feet. What is the distance diagonally across the park? City Parks ANSWER The distance diagonally across the park is about 212 ft

4 EXAMPLE 2 Using a 30 o –60 o –90 o Triangle Find the value of each variable in the triangle. Give the exact answer. You need to find the length of the shorter leg first in order to find the length of the longer leg. STEP 1 Find the length of the shorter leg. hypotenuse = 2 shorter leg = 10 2 x = 5 x Rule for 30 o –60 o –90 o triangle Substitute. Divide each side by 2.

5 EXAMPLE 2 STEP 2 Find the length of the longer leg. Rule for 30 o –60 o –90 o triangle Substitute. 3 longer leg = shorter leg = 3 y 5 The shorter leg is 5 units long. The longer leg is 5 units long. 3 ANSWER Using a 30 o –60 o –90 Triangle

6 GUIDED PRACTICE for Example 2 Find the value of each variable. Give the exact answer. 2. ANSWER The value for x = 25 in. y = in. 25 3

7 GUIDED PRACTICE for Example 2 3. ANSWER The value for x = 7m, y = 14m

8 GUIDED PRACTICE for Example 2 4. ANSWER The value for x = 2cm and y = 2 cm 3

9 EXAMPLE 3 Rotating 180 The pyramid ski show is a common attraction at water parks. Find the horizontal distance from the pyramid to the boat. Water Ski Show STEP 1 Find the length of the shorter leg. hypotenuse = 2 shorter leg = 26 2 x = 13 x

10 EXAMPLE 3 STEP 2 Find the length of the longer leg. longer leg = shorter leg 3 The horizontal distance is about 23 feet. ANSWER = 3 y 13 22.52 Using a Special Right Triangle

11 GUIDED PRACTICE for Example 3 5. What If? 3 In Example 3, suppose that the horizontal distance from the pyramid to the boat is 15 feet. What is the distance from the top of the pyramid to the boat? ANSWER The distance from the top of the pyramid to the boat is 30 ft.


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