Download Solving Right Triangles

The Mathematics 11
Competency Test
Solving Right Triangles
As mentioned already, to solve a right triangle means to calculate all unknown lengths and
angles.
Now, every triangle (including right triangles) have three angles and three sides, for a total of six
parts (as they are called). To solve a triangle, you need to know three of the six parts, and the
three known parts must include the length of at least one side.
Since we always know at least one angle of a right triangle (the right angle itself), then the
minimum additional information needed to solve a right triangle is either
(i) the length of two of the three sides (called the ‘ss’ case)
or
(ii) the length of one side and the value of one of the acute angles (called the ‘sa’ case)
Some books give formal strategies for each type of right triangle problem. Here, we just advise
the following procedure:
(i) list the values of the parts that are known (you don’t need to list the right angle)
(ii) list the symbols for the parts which are unknown and must be calculated.
(iii) then, use Pythagoras’s Theorem and/or the definitions of the principal trigonometric functions
and the inverse trigonometric functions to calculate each of the unknown parts, one-by-one. It is
a good strategy to set up your calculations so that they use just the original known values given in
the statement of the problem. Then if you make an arithmetic error, it doesn’t affect other
calculations.
B
Example 1: Solve the right triangle shown in
the sketch.
solution:
We are given
31
42
a = 31 and c = 42
and so we must determine b, A, and B.
When two sides of a right triangle are given,
the third side can always be determined using
Pythagoras’s Theorem:
A
b2 = c 2 – a 2
= 422 – 312 = 803.
Thus
b=
803 ≅ 28.34
rounded to two decimal places. Then
David W. Sabo (2003)
Solving Right Triangles
Page 1 of 4
sin A =
31
,
42
so
 31 
0
A = sin−1 
 ≅ 47.57
42


Finally,
cos B =
31
42
and so
 31 
B = cos−1   ≅ 42.430
 42 
Thus the required solution is
b ≅ 28.34, A ≅ 47.570, and B ≅ 42.430.
Notice how we organized our work in this example so that all calculations were based only on
information given in the original problem. None of the calculations involved numbers we had
previously calculated ourselves. This is a good strategy, because then if we had made an
arithmetic error in one step of the solution, it wouldn’t affect the correctness of results of other
calculations. This sort of defensive strategy is always possible when solving right triangles.
Example 2: Solve the right triangle sketched to
the right.
B
solution:
We are given
A = 270 and b = 412 m
To solve this triangle, we need to compute values
for B, a, and c.
270
When one of the acute angles is given, the
easiest way to compute the other is by
subtraction from 900:
A + B = 900
So
A
(for a right triangle)
270 + B = 900
giving
B = 900 – 270 = 630.
David W. Sabo (2003)
Solving Right Triangles
Page 2 of 4
Then, for side a, we have
tan A =
a
b
or
tan270 =
a
412 m
which gives
(
)
a = ( 412 m ) tan270 ≅ 209.92 m .
Finally, for side c,
cos A =
b
c
or
cos 270 =
412 m
c
giving
c=
412 m
≅ 462.40 m
cos 270
Thus, the required solution is:
B = 630, a ≅ 209.92 m, c ≅ 462.40 m.
Again, note the strategy employed in the previous example. To calculate ‘a’, we looked for a
trigonometric ratio that involved A and b (our given quantities) and ‘a’ (the unknown we wished to
determine). Similarly, to determine c, we looked for a trigonometric ratio involving A, b, and c. In
both of these cases this meant that the definition of the selected trigonometric ratio would amount
to an equation with one unknown, which
was then easily solved.
Example 3: Solve the right triangle
shown in the figure to the right.
A
520
solution:
The solution here is that there is no
solution. Knowing all three angles of a
right triangle is not enough information to
be able to calculate the lengths of any of
the three sides. In fact, there is an infinite
number of right triangles that have these
three angles – each of a different size.
David W. Sabo (2003)
380
B
Solving Right Triangles
Page 3 of 4
So, we cannot obtain a unique solution for this triangle.
Example 4: Solve the right triangle with c = 26.8 cm and B = 37.560.
solution:
It is probably best to start by making a
sketch, as is done to the right.
A
In addition to the right angle, of course, the
known parts of the triangle, B and c, are
given in the statement of the problem. So,
we need to determine the values of A, a,
and b.
b
First,
A + B = 90
so
37.560
0
0
B
A + 37.56 = 90
a
C
0
giving
A = 900 – 37.560 = 52.440
Then
sin B =
b
c
gives
sin37.560 =
or
b
26.8 cm
b = (26.8 cm)(sin 37.560) ≅ 16.34 cm
Finally, using
cos B =
a
c
or
cos37.560 =
a
26.8 cm
gives us
a = (26.8 cm)(cos 37.560) ≅ 21.24 cm
Thus, the required solution is;
A = 52.440, a ≅ 21.24 cm, b ≅ 16.34 cm.
David W. Sabo (2003)
Solving Right Triangles
Page 4 of 4