4.1 APPLICATIONS INVOLVING RIGHT TRIANGLES
If c2 = a2 + b2
then we know this is a right
triangle and therefore
α
c
b
α + β = 90°
β
a
Example 1 Solving a Right Triangle
Solving a Right Triangle
If b=2 and α = 40°, find a,c, and β
c
We now right away that β=50° because
it’s a right triangle and the other acute
angle is 40°.
40°
a
2
2 tan 40° = a = 1.68
tan 40° =
b=2
(adj to 40°)
β
a (opp to 40°)
Now that we know a and b we can use
the Pythagorean Theorem to solve for c,
or we can use sec 40 °.
c = a 2 + b 2 = (1.68) 2 + 2 2 ≈ 2.61
or
sec 40° = hyp / adj = c / b = c / 2
c = 2 sec 40° = 2.61
PDF created with pdfFactory Pro trial version www.pdffactory.com
Example 2
α
b
If a=3 and b=2, find c, and angles α and β
Use Pythagorean Theorem to find c:
c = a 2 + b 2 = 32 + 22 = 13 ≈ 3.61
(adj to
α)
c
β
a (opp to α)
Now that we have all 3 sides, a, b, and c, we can find any trig
function and take inverse to find the angles.
tan α= opp/adj = 3/2 = 1.5
α = tan-11.5 ≈ 56.3°
Since α+β = 90°, then β = 90°-56.3°= 33.7°
NOW YOU DO #9
Example 3 Finding the Inclination of a Mountain Trail
What is the inclination (or grade) of the trail?
The grade is the angle that measures how steep the hill is.
Use Pythagorean theorem to find a trig function for the inclination, θ
We know the hypotenuse is 14,100. The opposite leg, is
11,100 – 8,000 = 3,000.
sin θ = 3,100/14,100
Use sin-1 to find θ
=
T ra i l
1 4 ,1
et l
0 0 fe
ong
θ
Elev.
8000 ft
above sea level
NOW YOU DO #25
PDF created with pdfFactory Pro trial version www.pdffactory.com
Elev.
11,100 ft
above sea level
Example 7 Gibb’s Hill Lighthouse, Revisited
Horizon Limit
39
60
36
2/5
28
0+
Radius of Earth =
3960 miles
s
θ
36
2f
t.
In Week 1, we calculated the line-ofsight distance from the lighthouse to the
horizon at the direct distance from the
top of the lighthouse to the horizon.
Now let’s calculate the distance in
nautical miles, which is measure of the
angle θ in minutes.
3960
≈ 0.999982687
 362

+ 3960 

 5280

Change calculator mode to degrees
cos θ =
er
nt rth
e
C Ea
of
θ = cos −1 (0.999982687) ≈ 0.33715°
Use DMS function on calcutor to
convert to minutes.
θ ≈ 20.23' = 20.23 nautical miles
The distance in statute miles, s, is the actual arc length of the earth from one
point to another.
Arc length = s = rθ where θ is in radians.
Change calculator mode back to radians.
θ = cos-1(0.999982687) ≈ 0.00588 radian
r is the radius of the earth = 3960 miles.
s = 3960(0.00588) = 23.3 miles
Notice that this is the same answer that we got for the direct line-of-sight
distance in Project 1. That is because the height of the lighthouse is so
miniscule compared to the radius of the earth. Our diagram is definitely not
to scale!
PDF created with pdfFactory Pro trial version www.pdffactory.com
Example 9
A Boeing 777 aircraft takes off on “Runway 2 LEFT” which has a bearing of N20°E
After flying for 1 miles, the pilot of the aircraft requests permission to turn 90 degrees
and head toward the northwest. After the plane goes 2 miles in this direction, what
bearing should the control tower use to locate the aircraft?
N
β
θ
20°
1m
β=Bearing
= θ- 20°
Initial bearing of N20°E means the plane is headed
in the direction 20° East of North.
iles
ile
2m
W
E
Since the plane turned 90°, we have a right triangle with
one leg being the 1 mile it started out with, and the other
leg being the 2 miles it went after turning 90°.
2
=2
1
θ = tan −1 2 ≈ 63.4°
tan θ =
S
Bearing = β = θ − 20° = 63.4° − 20° = N 43.4°W
HOMEWORK
p. 293 #19-35 ODD
PDF created with pdfFactory Pro trial version www.pdffactory.com