Transcript Right Triangle Trigonometry - FIT

Right Triangle
Trigonometry
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
sinθ=
opposite
x
sin27°=
1.45m
X = 2.85 m
substitute.
x
sin27°(x)= 1.45m
sin 27°
Sine theta is equal to
opposite over hypotenuse.
Multiply sin27° and
hypotenuse.
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 32 o.
Find the length from Rey’s height if the
overall length of
their shadow is
6.49 m.
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
Rey
x
Shayne
1.78m
32°
6.49 m
y
Tan θ=
opposite
adjacent
Tan 32°=
1.78km
Tan θ is equal to opposite
over adjacent.
substitute.
x
Tan 32°(x)= 1.78 km
Tan 32°
Tan 32°
Cross multiply. divide tan
32° to the opposite side.
answer:
y = 2.85 km
y equals 2.85 km
Using proportion property:
The height of the shadow
casted by Rey is:
Height = 6.49m – 2.85m
= 3.64m
x
1.78m
=
3.64m
2.85m
(2.85m) x = (1.78m)(3.64m)
2.85 m
X = 2.27 m
Final answer:
the height of Rey
is 2.27 m.
Submitted by:
Rey PJ Enerio
Rekkah Niña Buliga
Veronica Shayne Francisco
Sheila Mae Gabe
Irene Pilar Tecson