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Solve two right triangle trigonometry problems involving height calculations using given angles and lengths.
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Problem 1: • Mimi was walking on the school ground late in the afternoon. While walking, she was not looking on her steps. Suddenly, she stepped on a banana peel.
Luckily, Rekkah and Irene, her classmates were there to catch her from falling on the ground.
A scalene triangle was formed between her and Irene. A right angle was formed below Irene’s arms and an angle measuring 27o from Mimi’s armpit. Irene has a height of 1.45 m excluding her head. Find the length of Mimi’s body when she reached her slanting position. 27° 1.45 m x
Solution: Let x be the hypotenuse Given the angle of 27° and an opposite side of 1.45 m. 27° 1.45 m x
opposite • Sine theta is equal to opposite over hypotenuse. sinθ= x 1.45m • substitute. sin27°= x • Multiply sin27° and hypotenuse. sin27°(x)= 1.45m sin 27° X = 2.85 m Final answer: x equals 2.85 m.
Problem 2: It was past afternoon that time. Shayne and Rey was about to go home. Waving goodbye, Shayne, with a height of 1.78m, casted a shadow with an angle of elevation of 32o. Find the length from Rey’s height if the overall length of their shadow is 6.49 m.
Rey Shayne Solution: let y be the length of the shadow of Shayne let x be the length of the height of Rey given 1.78m height of Shayne 32° angle of elevation 6.49 m total shadow casted x 1.78m 32° 6.49 m y
opposite • Tan θis equal to opposite over adjacent. Tanθ= adjacent 1.78km Tan32°= • substitute. x • Cross multiply. divide tan 32° to the opposite side. Tan32°(x)= Tan 32° 1.78 km Tan 32° answer: y equals 2.85 km y = 2.85 km
Using proportion property: The height of the shadow casted by Rey is: Height = 6.49m – 2.85m = 3.64m 1.78m 2.85m x 3.64m Final answer: the height of Rey is 2.27 m. = (2.85m) x = (1.78m)(3.64m) 2.85 m X = 2.27 m
Submitted by: Rey PJ Enerio Rekkah Niña Buliga Veronica Shayne Francisco Sheila Mae Gabe Irene Pilar Tecson