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Trigonometry
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Solving Right Triangles
Solving Right Triangles
This is an important milestone in our trigonometric journey. Solving triangles is one if not the central question in
trigonometry. We have already had some success at this using, pythagoras, the 30-60 Theorem, 45-45 Theorems,
the almost-triangles, and the 23-67 Theorem.
We are now ready to fulfill one of the great trigonometric promises, to measure buildings, trees, mountains, starts,
planets, anything by just looking at it... and measuring an angle or twoa . This can be done in many ways. We
will illustrate two, one with just common sense and another solving equations. The general idea is to compare a
known side with one that is unknown. The unknown side can be determined so long as the ration is known. For
example, recall that on a 30-60 Triangle, the small to hypothenuse ratio is 50%, and suppose the hypothenuse is
known to be 14 ft, then since the ratio is 50% the small one must be 7 feet. We are now ready to repeat this sort
of reasoning, not just for 30-60 or 45-45 but for all right triangles.
Example 1
Example 2
The known side is the hypothenuse, 14ft., the unknow side is the opposite of 30◦ , s. The function
that describes the opposite to hypothnuse ratio is
the sine function. It says that the sin 30◦ = .5, i.e.
the opposite to hyp ratio is 50%, thus the opposite
is 50% of 14 ft., which is 7ft. Using the equations
the solution would look something like this:
sin 30◦
14ft
15
◦
s
.5
=
=
.5(14 f t) =
7 ft =
s
14 f t
s
14 f t
s
The known side is the side adjacent to the angle
15◦ , 8 ft., the unknow side is the opposite side,
s. The function that describes the opposite to adjacent ratio is the tangent function. The tangent
function says that the opposite to adjacent ration
for such a triangle is tan 15◦ ≈ .278 = 27.8%, thus
the opposite is about 27.8% of 8 ft., which is approximately 2.2 ft. Using the equations the solution would look something like this:
tan 15◦
=
s
15
◦
.278 ≈
8 ft
.278(8 f t) ≈
s
2.2 f t ≈
Example 3
15
◦
cos 15◦
8
s
Example 4
The known side is adjacent to the angle 15◦ , 8 ft.,
the unknow side is the hypothenuse, r. The function that describes the ratio between adjacent and
hypothenuse is the cosine function. It says that
the cos 15◦ ≈ .966, i.e. the adjacent side is approximately 96.6% of the hypothenuse. Thus the
hypothenuse is 8ft divided by 96.6% which is approximately 8.3 ft. Using the equations the solution would look something like this:
r
s
8 ft
s
8 ft
s
=
.966 ≈
(.966)(s) ≈
s ≈
8
s
8
s
8
8(.966) = 8.3
This time we look at the same triangle as example
3, but from an diferent point of view. This time we
look at it from the other corner, the 75◦ -corner.
From here, the known side is the side opposite,
while the unknown is still the hypothenuse. The
sine function was born to describe such a ratio,
opp/hyp. Note we should expect the same answer
since we calculated it in example 3.
r
8 ft
75
◦
sin 75◦
r sin 75◦
8
r
= 8
=
r
=
r
≈
8
sin 75◦
8
= 8.3
.966
a some
artistic liberty was taken in making this statement
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2007
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c
2007
MathHands.com
Trigonometry
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1. Solve ALL sides of the right triangle.
y
r
y
r
4.5
39◦
5
25◦
2. Solve ALL sides of the right triangle.
r
y
r
7.
5
x
10◦
7
55◦
3. Solve ALL sides of the right triangle.
r
8.
5
x
20◦
y
r
8
15◦
4. Solve ALL sides of the right triangle.
r
9.
2
x
70◦
y
r
15
17◦
r
12
5. Solve ALL sides of the right triangle.
10.
y
r
80◦
x
37◦
10
6. Solve ALL sides of the right triangle.
r
11.
50◦
15
x
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2007
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Trigonometry
Sec.
MathHands.com
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r
12.
17
x
23◦
river
17◦
25◦
10 yd.
r
13.
49◦
8
16. (a) Determine the distance from the tree trunk to the
opposite side of the river as shown.
x
14. Just by looking at him, can you figure out how big Diego’s
head is? See picture for angles & measurements.
river
20◦
23◦
5 yd.
40◦
49◦
(b) (con’t from previous exercise) How far away is the
tree from the river? Use the diagram below.
2 in.
d
river
18
◦
15. Determine the distance from the tree trunk to the opposite side of the river as shown.
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2007
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