Download 7.1 Right Triangle Trigonometry: Applications

Section 7.1 Notes Page 1
7.1 Right Triangle Trigonometry: Applications
We will come back to right triangles again. In these questions, if it asks you to solve the triangle then that
means you need to find all sides and angles.
EXAMPLE: Solve the triangle:
B
35 o
A
200 ft
C
In this problem we need to find side AB, side AC, and m∠A (measurement of angle A). First we can find
m∠A . The sum of all the angles in a triangle is 180 degrees. We have a 35 degree angle and a 90 degree
angle, so m∠A = 180 − 35 − 90 = 55 o . Now we want to find side AB. Let’s call this length x. From our picture
above let’s use the 35 degree angle. The side opposite this angle is 200 ft. The hypotenuse is x, which is what
200
. We can solve for x by cross
we are trying to find. We can set up the following equation: sin 35 o =
x
200
multiplying. We get x =
= 348.69 ft. Now we can solve for side BC, which we are calling y. Again if
sin 35 o
we use the 35 degree angle, the y is the adjacent angle and 200 ft is still the opposite side. We will use the
200
. We can solve for y by doing cross multiplication:
following equation to solve for y: tan 35 o =
y
200
y=
= 285.63 ft. Now we know the measurement of all three sides and angles, so it is solved.
tan 35 o
EXAMPLE: Solve the triangle:
B
I will let side BC be x. Now we can use the Pythagorean Theorem to find
( )
2
find it: 3 2 + x 2 = 3 5 . Solving this will give us: 9 + x 2 = 45 , so
3 5
A
3 cm
C
x = 6 . To find m∠A , we can set up the following trig equation:
3
. So we have cos A = 0.4472 . We need to take the inverse
cos A =
3 5
cosine to get our answer. So A = cos −1 0.4472 = 63.44 o . To find m∠B
we will subtract 90 degrees and 63.44 degrees from 180 degrees. We will
get: m∠B = 180 − 63.44 − 90 = 26.56 o . Now our triangle is solved.
Section 7.1 Notes Page 2
EXAMPLE: A security camera in a neighborhood bank is mounted on a wall 11 ft above the floor. What angle
of depression should be used if the camera is to be directed to a spot 6 ft above the floor and 12 ft from the
wall?
We first need to draw a picture for this one:
A
5 ft
A
12 ft
6 ft
We need to find angle A which is the angle of depression. This is the same as the angle A inside the triangle.
When looking at the angle A inside the triangle we see that the opposite side is 5ft and the adjacent is 12 feet.
5
We can set up the following equation: tan A = . This can also be written as tan A = 0.4167 . In order to find
12
A we need the inverse tangent: A = tan −1 0.4167 = 22.62 o . Therefore the angle of depression should be 22.62 o
EXAMPLE: The eyes of a basketball player are 6 ft above the floor. The player is at the free-throw line, which
is 15 ft from the center of the basket rim. What is the angle of elevation from the player’s eyes to the center of
the rim if the rim is 10ft above the floor?
We first need to draw a picture for this one:
A
4 ft
A
15 ft
6 ft
We need to find angle A which is the angle of depression. This is the same as the angle A inside the triangle.
When looking at the angle A inside the triangle we see that the opposite side is 4ft and the adjacent is 15 feet.
4
We can set up the following equation: tan A = . This can also be written as tan A = 0.2667 . In order to find
15
−1
A we need the inverse tangent: A = tan 0.2667 = 14.93 o . Therefore the angle of depression should be 14.93 o .
Bearing
Bearing is a way to measure direction. There are three parts to bearing. The
first part is either a N or S. The second part is an acute angle that is ALWAYS
measured from a vertical axis. The third part is an E or W. The picture to
the right shows how North, South, East, and West is orientated. The next
few examples show how to draw bearings.
Section 7.1 Notes Page 3
o
EXAMPLE: Draw the following: N 40 E .
North is always straight up. First we go straight north. Then is says to go 40 degrees to the east. So from the
north we will measure 40 degrees. Our picture is:
EXAMPLE: Draw the following: N 65 o W .
North is always straight up. First we go straight north. Then is says to go 65 degrees to the west. So from the
north we will measure 65 degrees. Our picture is:
EXAMPLE: Draw the following: S 70 o E .
South is always straight down. First we go straight south. Then is says to go 65 degrees to the east. So from
the south we will measure 70 degrees. Our picture is:
Section 7.1 Notes Page 4
EXAMPLE: An airplane takes off from a runway at a bearing of N 20 E . After flying 1 mile in this direction
the plane makes a 90 degree turn to the left. After flying 2 more miles in this new direction, what is the bearing
form the airport to the plane?
N
2 miles
First we need to draw a picture as shown. Notice the 20 degrees is
measured from the north since this is a bearing. You will end up
in the second quadrant.
20 o 1 mile
W
E
A
We need to find the measurement of angle A. Then we will subtract
20 degrees to find the bearing in the second quadrant. In order to find
2
angle A we can use the formula: tan A = . Again we need to use the
1
−1
inverse tangent to find angle A: A = tan 2 = 63.43 o . This is the entire
S
o
o
o
angle of A, so we need to subtract 20: 63.43 − 20 = 43.43 . We need
to write this as a bearing, so our answer is: N 43.43 o W .
o
EXAMPLE: A ship leaves the port of Miami with a bearing of S 80 o E and a speed of 15mph. After 1 hour the
ship turns 90 degrees towards the south. After 2.5 hours maintaining the same speed, what is the bearing from
the port to the ship?
First we draw the picture. The 80 degrees is measured from
the south. After 1 hour the ship will travel 15 miles. After
2.5 hours the boat travels (2.5)(15) = 37.5 miles as indicated
on the diagram. If we can find x, then we can subtract this
from 80 degrees to get the bearing measured from the south
axis. When looking at x, the opposite side is 37.5 and the
adjacent side is 15, so we need to use tangent again. So
37.5
. This means tan x = 2.5 . To find x we need to
tan x =
15
take the inverse tangent of both sides: x = tan −1 2.5 = 68.2 o .
This is x, but in order to find the bearing we need to subtract
this from 80 degrees since bearings are always measured from
the vertical. 80 o − 68.2 o = 11.8 o . Finally we need to write our
bearing, which is S11.8 o E .