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- The Adventure of Rodger and Earl
- I had the best day!
- Me too!!
- just keep swimming, just keep swimming
- Meet Rodger (left) and Earl (right)!
- C'mon Bro, just jump!
- After a long day of playing on the beach, Rodger and Earl have to get home to dinner. They will need to use trig to plan around obstacles that get in their way.
- Yes, it absolutely is
- Is this necessary? You already have one bone
- Rodger and Earl must swim part way around the base of a semi-circle rock to continue their journey home. Earl thinks the rock is 15 ft wide and the path of least resistance is 45 degrees or π/4 radians from their current location. Using the arc length formula, Rodger uses the arc length formula to calculate π/4 (15/2) = 5.89 ft. They must swim about 6 feet to get to the other side of the formation.
- I'm so tired
- Finally.
- Now, Rodger and Earl must jump to the rope and swing to cross the river. Earl has made the jump but Rodger is a thinker. He wants to know the angle at which to jump. To find this, Rodger uses SOH of SOH-CAH-TOA. He estimates the distance to the rope is 8ft and the distance to under the rope is 6ft.  He calculates sin-1 (6/8) to jump at an angle of 49 degrees.
- The pups are on the home stretch but suddenly Earl stops for the scent of bones. He must dig to see if there are any more bones. The space they have to dig up is a triangle. Using Heron's formula, Rodger calculates the area. ___________________________________. The area they must dig is 5.3 ft^2.
- WOOOOHOOO!! Rodger and Earl made it home!
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