Download Solving Right Triangle Problems

The Mathematics 11
Competency Test
Solving Right Triangle Problems
Right triangle problems are problems requiring the determination of some geometric property or
properties, and the solution of the problem requires the solving of a right triangle. We’ve already
illustrated strategies for solving right triangles. The main new feature in right triangle problem
then is to recognize how the solution of the
problem itself can be achieved through
N
solving an appropriate right triangle.
E
W
Example 1: Hank needs to determine the
distance across a wide river in a north-south
direction between points A and B, as indicated
in the sketch to the right. He locates point C,
which is 215 meters due west of point A, and
is able to measure that from point C, point B
makes an angle of 36.40 north of the line from
C to A. Determine the distance AB.
S
river
C
B
river
36.40
solution:
A
The geometry of the situation is summarized
in the second figure to the right. The points A, B,
and C form a triangle. Since side AB is in a northB
south direction, and side CA is in an east-west
direction, the angle at point A must be a right angle.
(The labelling of the vertices of this right triangle is
not exactly according to the usual convention, but
that’s because the points forming the vertices were
labelled before it was known that they would be
36.40
forming the vertices of a right triangle.) Further, to
A
C
215 m
say that the side CB is 36.40 north of east means
that the angle of the right triangle at point C is 36.40.
This gives us enough information to solve the right triangle, and in particular, to determine the
length of side AB.
We have
tan36.40 =
AB
AB
=
AC 215 m
where we are using the notation AB and AC to indicate the lengths of the sides between these
pairs of points, respectively. Therefore
AB = (215 m)(tan 36.40) ≅ 158.5 m
rounded to one decimal place.
Thus, the required distance across the river between points A and B is 158.5 m.
David W. Sabo (2003)
Solving Right Triangle Problems
Page 1 of 7
Example 2: A vertical television transmission tower is
112 m high. Support cables attached to its top are to be
anchored in the ground at points located so that the
support wire makes an angle of 750 with the ground.
Assuming the ground is horizontal in the vicinity of the
base of the tower and that the support cables are
stretched tight enough so that they essentially form a
straight line, determine how far the cable anchors will be
from the base of the tower, and how long the cables must
be.
tower
solution:
Since the tower is vertical and the ground is horizontal, the
ground, tower, and cable form a right triangle as sketched in the
second figure to the right. We are given the length of one side of that
right triangle, and the measure of one of the acute angles. The
problem asks us to determine the lengths of the other two sides,
labelled x and y in the sketch.
750
y
From this diagram
sin 750 =
112 m
y
750
x
and so
y=
112 m
≅ 115.95 m
sin 750
Also,
tan 750 =
112 m
x
and so
x=
112 m
≅ 30.01 m
tan 750
Thus, the support cables must be 115.95 m long, and they must be anchored in the ground at a
distance of 30.01 m away from the base of the tower.
Example 3: The sketch to the right shows
the cross section of a roof of a fairly large
house. One side must have an angle of 200
above the horizontal in order to facilitate the
installation of solar electricity batteries.
However, the peak of the roof may be only
3.50 m above the base level. Determine the
two lengths labelled y and z in the sketch,
David W. Sabo (2003)
z
y
θ
200
Solving Right Triangle Problems
14.50 m
Page 2 of 7
and also, determine whether the angle θ is greater than 450.
solution:
This seems like quite a complicated
problem, so we’ll have to be fairly
systematic in our work. The first thing to
recognize is that the cross-section of this
roof really consists of two right triangles,
sharing as a common side the vertical
line through the peak of the roof.
θ
200
We have enough information to solve the
right triangle on the left and so determine
the value of y. Since
sin 200 =
z
y
u
v
14.50 m
3.50 m
y
we get
y=
3.50 m
≅ 10.23 m
sin 200
At present, we do not have enough information to solve the right triangle on the right, of which z
and θ are parts, since we know the length of just one side in addition to the right angle. However,
we can determine the length of the side labelled v in the sketch as follows. For the triangle on the
left,
tan 200 =
3.50 m
u
giving
u=
3.50 m
≅ 9.62 m
tan 200
But
u + v = 14.50 m
so that
v = 14.50 m − u = 14.50 m − 9.62 m = 4.88 m
Now we have enough information to solve the triangle on the right. By Pythagoras’s Theorem,
z 2 = ( 3.50 m ) + ( 4.88 m ) = 36.0644 m 2
2
2
so that
z=
36.0644 m 2 ≅ 6.01 m
David W. Sabo (2003)
Solving Right Triangle Problems
Page 3 of 7
Also,
tanθ =
3.50 m 3.50 m
=
v
4.88 m
from which we can write
 3.50 
0
 ≅ 35.65
4.88


θ = tan−1 
We are now able to state a final solution to this problem. The length y is 10.23 m and the length z
is 6.01 m. The angle θ is 35.650, and so it is not greater than 450.
Example 4: A rectangular field measures 108 m
wide by 176 m long. If paths are constructed
joining opposite corners of the field, what angles
will they form where they cross?
176 m
solution:
We see from the sketch that two different angles
are formed where the diagonal lines intersect.
However, obviously
ϕ
θ
θ + ϕ = 1800
so if we can calculate either of the two required angles, we can get the other one by simple
subtraction from 1800.
There are various ways to solve this problem.
One way is as follows. Draw a vertical and
horizontal line through the point where the
diagonals intersect (the dotted lines in the second
sketch). These lines form a number of right
angles where vertical and horizontal lines meet in
the sketch, and hence a number of right triangles
appear. Now, since opposite sides of a rectangle
are parallel, and the diagonal lines form
transversals of these parallel sides here, it is easy
to demonstrate that
F
E
B
x x
A
C
D
∆ABC ∼ ∆DBC
and further, since they share the side BC, these two triangles are identical. (Technically, this
means that not only is ∆ABC similar to ∆DBC, but ∆ABC is said to be congruent to ∆DBC – they
have the same shape and the same size.) This means that
AC = DC =
176 m
= 88 m
2
and
David W. Sabo (2003)
Solving Right Triangle Problems
Page 4 of 7
BC = EA = EF =
108 m
= 54 m
2
But then
tan x =
AC 88 m
=
BC 54 m
and so
 88 
x = tan−1   ≅ 58.470
 54 
But
θ = x + x = 2x = 2(58.470) = 116.940
But then,
ϕ = 1800 - θ = 1800 – 116.940 = 63.060.
Thus, the diagonals of this rectangular field intersect forming angles of 116.940 and 63.060.
Example 5: A regular pentagon is inscribed in a circle of radius
5 m. Determine the area of the pentagon.
solution:
Pentagons seem to be a bit of a jump from triangles. However,
notice that if we draw lines joining the center of the circle to the
vertices of the pentagon, we end up with five triangles. In fact,
since this is a “regular pentagon,” all five triangles formed in this
way are identical. (By “regular” pentagon, is meant that all five
sides of the pentagon are of equal length, and all five
angles formed where the sides meet are equal. Since
each of the triangles formed by joining the center of the
circle to the vertices of the pentagon have two sides which
are equal to the radius of the circle, and a third side which
is equal in every triangle, you can easily use principles
discussed in connection with our study of similar triangles
to conclude that all five triangles are identical in size and
shape.)
θ
θ
θ
θ
θ
If we can calculate the area of one of these triangles, then
the area of the entire pentagon can be obtained by simply
multiplying by five.
Since all five triangles are identical, the angles, labelled θ in the diagram, that each forms at the
center of the circle must be the same in all five triangles. Since one complete rotation about the
center amounts to 3600, we must have
5θ = 3600
so
David W. Sabo (2003)
Solving Right Triangle Problems
Page 5 of 7
3600
θ=
= 720
5
As mentioned, the sides of each triangle which are
formed by the radial lines from the center of the circle to
the vertices of the pentagon at the circle itself, are all
equal to the radius of the circle, 5 m. Thus, each of the
five identical triangles making up the pentagon look like
the one in the sketch to the right. Since the two vertices
not equal to θ are opposite sides of equal lengths, those
two angles themselves must be equal. To show this,
we’ve labelled both of them ϕ.
This is not yet a right triangle, but it is easy to come up
with right triangles. Draw a line from the top vertex to
the opposite side of this triangle so that it divides the
top vertex into two equal angles of 360 each (as shown
in the second figure to the right). You should be able
to demonstrate that ∆ACD is similar to ∆BCD. (In fact,
again, they are not only similar, but identical in size
and hence congruent.) This means that ∠ADC =
∠BDC. But, since these two angles add up to a
straight angle (or 1800), they must each equal 900, so
they are right angles. Thus, ∆ACD and ∆BCD are right
triangles.
5m
720
5m
ϕ
ϕ
C
360 360
5m
5m
ϕ
ϕ
A
B
D
Now, for ∆ACD, we have
cos 360 =
C
h
5m
360 360
5m
5m
so that
h
h = (cos 360)(5 m) ≅ 4.045 m
ϕ
A
Also
1
b
sin 360 = 2
5m
ϕ
D
b
B
so that
1
b = ( sin 360 ) ( 5 m )
2
or
b = (2)(sin 360)(5 m) ≅ 5.878 m
Thus, the area of this small triangle, ∆ABC, is
David W. Sabo (2003)
Solving Right Triangle Problems
Page 6 of 7
1
1
bh = ( 5.878 m )( 4.045 m ) ≅ 11.8883 m 2
2
2
Finally, the area of the entire pentagon is five times this area, for a total of 59.442 m2.
This last example is considerably more complicated than any that you will encounter on the BCIT
Mathematics 11 Competency Test. However, it is worth studying because its solution involves
quite a number of the geometric and trigonometric principles that we have described earlier in
various parts of these notes.
David W. Sabo (2003)
Solving Right Triangle Problems
Page 7 of 7