Download 2. Directions and Bearings

Further Trigonometry
1.
Composite Figures
A diagram consisting of more than one triangle is said to be a composite figure.
For trigonometric problems involving a composite figure, first decide whether to use
sine, cosine or tangent, and then calculate the required length or angle.
Example 1
In the given diagram, find:
a. x
b. y
Solution:
Using a Construction Line
To solve some trigonometric problems, we need to convert the given triangle into two
right-angled triangles by drawing a perpendicular construction line from the vertex to
the opposite side.
Example 2
Find BC in the given diagram, rounded to 2 decimal places.
Solution:
Draw BD perpendicular to AC. Let BD = x cm, BC = y cm.
2. Directions and Bearings
The direction to a point is stated as the number of degrees east or west of north or south.
For example, the direction of A from O is N30ºE.
B is N60ºW from O.
C is S70ºE from O.
D is S80ºW from O.
Note:
N30ºE means the direction is 30º east of north.
The bearing to a point is the angle measured in a clockwise direction from the north
line.
For example, the bearing of P from O is 065º.
The bearing of Q from O is 300º.
Note:
The direction of P from O is N65ºE.
The direction of Q from O is N60ºW.
A bearing is used to represent the direction of one point relative to another point.
For example, the bearing of A from B is 065º.
The bearing of B from A is 245º.
Note:


Three figures are used to give bearings.
All bearings are measured in a horizontal plane.
Example 3
A boat sails from a certain port in the direction N30ºW. After the boat has sailed 20
km, how far is it west of the port?
Solution:
Let the boat be x km west of the port.
So, the boat is 10 km west of the port.
Example 4
A cyclist travels 10 km south, then 8 km east. Find the cyclist's bearing from her
starting point to the nearest degree.
Solution:
So, the cyclist's bearing is 141º from her starting point.
3. Angles of Elevation and Depression
The angle of elevation of an object as seen by an observer is the angle between the
horizontal and the line from the object to the observer's eye (the line of sight).
If the object is below the level of the observer, then the angle between the horizontal
and the observer's line of sight is called the angle of depression.
Example 5
From the top of a vertical cliff 40 m high, the angle of depression of an object that is
level with the base of the cliff is 34º. How far is the object from the base of the cliff?
Solution:
Let x m be the distance of the object from the base of the cliff.
So, the object is 59.30 m from the base of the cliff.
4. Three-Dimensional Problems
To solve a three-dimensional problem, it is important to be able to visualise right
triangles contained in a diagram. Then redraw the right triangles in two dimensions
and use an appropriate trigonometric ratio and/or apply Pythagoras' Theorem to obtain
the answer.
Example 6
Find the angle between the body diagonal and the base of a cube of side-length 5 cm.
Solution:
By Pythagoras' Theorem and triangle
ACD,
Taking the square root of both sides
gives
So, the angle between the body diagonal and the base of the cube is 35º16/.
5. Some formulas:
Sum formulas
cos(u - v)
We prove this formula using the concept of dot product of two vectors. (See theory about vectors)
With u corresponds one point p(cos(u),sin(u)) on the unit circle
With v corresponds one point q(cos(v),sin(v)) on the unit circle
The angle, corresponding with the arc qp of the circle, has a value u - v .
Now : p.q = 1.1.cos(u-v) .
But using the coordinates we also have p.q = cos(u).cos(v)+sin(u).sin(v).
Hence,
cos(u-v) = cos(u).cos(v)+sin(u).sin(v)
cos(u + v)
cos(u + v) = cos(u - (-v)) = cos(u).cos(-v)+sin(u).sin(-v)
cos(u + v) = cos(u).cos(v)-sin(u).sin(v)
sin(u - v)
sin(u - v) = cos(pi/2-(u-v)) = cos( (pi/2-u) +v )
= cos(pi/2 - u).cos(v)-sin(pi/2 - u).sin(v)
sin(u - v) = sin(u).cos(v)-cos(u).sin(v)
sin(u + v)
sin(u + v) = cos(pi/2-(u+v)) = cos( (pi/2-u) -v )
= cos(pi/2 - u).cos(v)+sin(pi/2 - u).sin(v)
sin(u + v) = sin(u).cos(v)+cos(u).sin(v)
tan(u + v)
sin(u + v)
sin(u).cos(v)+cos(u).sin(v)
tan(u+v) = ------------ = --------------------------cos(u + v)
cos(u).cos(v)-sin(u).sin(v)
Dividing the dominator and denominator by cos(u).cos(v) we have
tan(u) + tan(v)
tan(u+v) = ----------------1 - tan(u).tan(v)
tan(u - v)
In the same way, we have
tan(u) - tan(v)
tan(u-v) = ----------------1 + tan(u).tan(v)
sin(2u)
sin(2u) = sin(u + u) = sin(u).cos(u)+cos(u).sin(u) = 2sin(u).cos(u)
sin(2u) = 2sin(u).cos(u)
cos(2u)
cos(2u) = cos(u+u) = cos(u).cos(u)-sin(u).sin(u) = cos2 (u) - sin2 (u)
cos(2u) = cos2 (u) - sin2 (u)
tan(2u)
tan(u) + tan(u)
tan(2u) = -----------------1 - tan(u).tan(u)
=
2 tan(u)
--------------1- tan(u)tan(u)
tan(2u)
2 tan(u)
= ----------1- tan2(u)
Carnot formulas
1 + cos(2u) =
1+cos2 (u)-sin2 (u) =
2 cos2 (u)
1 - cos(2u) =
1-cos2 (u)+sin2 (u) =
2 sin2 (u)
1 + cos(2u) = 2 cos2 (u)
1 - cos(2u) = 2 sin2 (u)
t-formulas
From the Carnot formulas we have
cos(2u)
=
2 cos2(u) -1
2
= ------------ - 1
1 + tan2 (u)
1 - tan2(u)
= ------------1 + tan2 (u)
We know:
2 tan(u)
tan(2u)= -------------
1 - tan2 (u)
Hence,
sin(2u) =
2 tan(u)
----------1 + tan2 (u)
Let t = tan(u) , then
1 - t2
cos(2u) = --------1 + t2
;
2t
sin(2u) = -------- ;
1 + t2
2t
tan(2u) = ------1 - t2
;
These 3 formulas are called the t-formulas.