Transcript The Navigation Problem Trigonometry Mr. Brennan The Navigation Problem General Case A plane flies for 2.25 hours (from an airport) at a speed of 240 km/hr A plane.

The Navigation Problem
Trigonometry
Mr. Brennan
The Navigation Problem
General Case
A plane flies for 2.25 hours
(from an airport)
at a speed of 240 km/hr
A plane flies for t1 hours
(from an airport)
at a speed of s1 km/hr
on a course of 203 degrees.
on a course of c1 degrees.
Then on a course of 300 degrees
the plane flies for 3.5 hours
at a speed of 300 km/hr.
Then on a course of c2 degrees
the plane flies for t2 hours
at a speed of s2 km/hr.
At this time, how far is the
plane from the airport?
At this time, how far is the
plane from the airport?
The Navigation Problem
A plane flies for t1 hours
(from an airport)
at a speed of s1 km/hr
A
c1°
on a course of c1 degrees.
Then on a course of c2 degrees
the plane flies for t2 hours
at a speed of s2 km/hr.
At this time, how far is the
plane from the airport?
c2°
The Navigation Problem
A plane flies for t1 hours
(from an airport)
at a speed of s1 km/hr
on a course of c1 degrees.
Note: Remember that the navigation problems do not measure
angles in standard position, but are oriented from North.
Whenever there angle measures you can draw in a North-South
line to help you orient the angle. Remember to draw a picture.
N
A
c1°
N
A plane flies for t1 hours
(from an airport)
at a speed of s1 km/hr
on a course of c1 degrees.
A
c1°
d1 = s1 * t1
At this time the plane has
traveled a distance of
d1 = s1 * t1 km
This distance becomes the first
side of a triangle
N
The Navigation Problem
N
A plane flies for t1 hours
(from an airport)
at a speed of s1 km/hr
A
c1°
on a course of c1 degrees.
d2 = s2 * t2 km
d1
c2°
Then on a course of c2 degrees
the plane flies for t2 hours
The distance covered by the second direction
at a speed of s2 km/hr.
d2 = s2 * t2 km
d2 becomes the second side of the triangle
The Navigation Problem
A plane flies for t1 hours
(from an airport)
at a speed of s1 km/hr
on a course of c1 degrees.
d3
A
c1°
d2
d1
c2°
B
Then on a course of c2 degrees
the plane flies for t2 hours
at a speed of s2 km/hr.
We want to find the distance d3,
which is the third side of the triangle.
At this time, how far is the
plane from the airport?
We can start by finding the measure of
angle B, then with sides d1 and d2 we an
use the law of cosines to find side d3.
N
The Navigation Problem
We want to find the distance d3,
which is the third side of the triangle.
N
d3
A
c1°
We can start by finding the measure of
angle B, then with sides d1 and d2 we an
use the law of cosines to find side d3.
d1
d2
(C1 - 180°)
(360 –c2°)
We need to find the measure of angle B. We can do
this by finding the sum of the two smaller angles
that make up angle B.
c2°
B
First – look at the shaded region from course C1.
The measure of the shaded region is C1 - 180°.
Because the lines going North are parallel, the right half of angle B also has a measure
of (C1 - 180°.) [Alternate Interior Angles are congruent. ]
As shown, the measure of the left side of angle B is (360 – c2)°
The measure of angle B = (360 – c2) + (c1 – 180)
= 180 + c1 – c2
N
The Navigation Problem
We want to find the distance d3,
which is the third side of the triangle.
We can start by finding the measure of
angle B, then with sides d1 and d2 we an
use the law of cosines to find side d3.
The measure of angle B = (360 – c2) + (c1 – 180)
= 180 + c1 – c2
d3
(d3)2 = (d1)2 + (d2)2 – 2(d1)(d2) cos (B)
So d3 =
(d1)2 + (d2)2 – 2(d1)(d2) cos (B)
A
c1°
d1
d2
(C1 - 180°)
c2°
The law of cosines says
N
B
(C1 - 180°)
The Navigation Problem
A plane flies for 2.25 hours
(from an airport)
at a speed of 240 km/hr
on a course of 203 degrees.
Then on a course of 300 degrees
the plane flies for 3.5 hours
at a speed of 300 km/hr.
At this time, how far is the
plane from the airport?
The Navigation Problem
A plane flies for 2.25 hours
(from an airport)
at a speed of 240 km/hr
A
c1°
on a course of 203 degrees.
d1
d2 = 1050 km
Then on a course of 300 degrees
the plane flies for 3.5 hours
at a speed of 300 km/hr.
At this time, how far is the
plane from the airport?
= 540 km
c2°
d1 = speed * time
240 km/hr * 2.25 hr = 540 km
The distance covered by the second direction
d2 = speed * time
d2 = s2 * t2
d2 = 300 km/hr * 3.5 hr = 1050 km
The Navigation Problem
A plane flies for 2.25 hours
(from an airport)
at a speed of 240 km/hr
A
c1°
on a course of 203 degrees.
d2 = 1050 km
d1
= 540 km
23°
60°
Then on a course of 300 degrees
the plane flies for 3.5 hours
at a speed of 300 km/hr.
c2°
Calculate the measure of angle B.
At this time, how far is the
plane from the airport?
B
From course1 the measure of the right half of
angle B = (203 – 180) = 23°
From course2 the measure of the left half of
angle B = (360 – 300) = 60°
The measure of angle B = 23° + 60° = 83°
The Navigation Problem
A plane flies for 2.25 hours
(from an airport)
at a speed of 240 km/hr
d3
A
c1°
on a course of 203 degrees.
d2 = 1050 km
d1
= 540 km
23°
60°
Then on a course of 300 degrees
the plane flies for 3.5 hours
at a speed of 300 km/hr.
At this time, how far is the
plane from the airport?
c2°
The law of cosines says
B = 83°
(d3)2 = (d1)2 + (d2)2 – 2(d1)(d2) cos (B)
So d3 =
(d1)2 + (d2)2 – 2(d1)(d2) cos (B)
D3 = 5402 + 10502 – 2(540)(1050) cos (83)
D3 =1120.67 km