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### Finding Sides and Angles

```
Date: 01/22/2001 at 01:22:36
From: john yee
Subject: Geometry

Dear Dr. Math,

I have a real-life question that goes beyond my high school training
or I've just simply forgotten how to do it. I suspect this is a

This is a right triangle of which one side and the right angle are
known. Also given is an elevation value that is not the true length
of any of the triangle's sides.

From a horizontal and vertical intersection begins a 5/32 long leg of
a right triangle 10 degrees upward from the horizontal line.  At the
distance of 5/32 another leg upward is perpendicular to the 5/32 line
and ends on a line parallel and 2 1/2 away from the horizontal line.
This point and the horizontal vertical intersection determine the
hypotenuse.

Can you determine the length of the hypotenuse and an acute interior
angle of this right triangle?

Can you show me how to do this and identify the subclass of geometry

Thanks again.
```

```
Date: 01/22/2001 at 10:52:00
From: Doctor Rob
Subject: Re: Geometry

Thanks for writing to Ask Dr. Math, John.

This is a problem in trigonometry. There is enough information to find
all the sides and angles of the triangle.  Here is a rough diagram:

C
---o---------------------------------- F horizontal line
|\ 80 degrees
|\\
|  \
| \ \
|   y\
|  \  \
|      \
|   \   \
|        \
|    \    \
|          \
|     \     \
|            \
|      \      \
|              \
|       \       \
|                \
|        \x       \
|5/2               \
|         \         \
|                    \
|          \          \
|                      \
|           \           \
|                        \
|            \            \
|                   5/32 _,o B
|             \      _,-'
|                _,-' 10 degrees
---o--------------o------------------ E   horizontal line
D              A

The triangle described is ABC, <B = 90 degrees, <BAE = 10 degrees.
Length AB = 5/32 and length CD = 5/2. Then <BCD = 10 degrees (since
the sides of <BCD are perpendicular to the sides of <BAE).

Let length AC, which you seek, be x, and <ACB = y. Then, looking at
right triangle ACD,

cos(10-y) = (5/2)/x = 5/(2*x),

and looking at right triangle ABC,

sin(y) = (5/32)/x = 5/(32*x).

This forms a system of two equations in two unknowns x and y, which
you need to solve.  The solution you seek will have x > 5/2 and
0 < y < 10 degrees.

I would solve this system by using the trigonometric identities

cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b),
sin^2(a) + cos^2(a) = 1.

The first equation above and the first identity give

cos(10)*cos(y) + sin(10)*sin(y) = 5/(2*x).

Substituting for sin(y) its equal from the second equation above gives

cos(10)*cos(y) + sin(10)*5/(32*x) = 5/(2*x).

Solving for cos(y), you get

cos(y) = [5/(2*x) - sin(10)*5/(32*x)]/cos(10),
= 5*[16*sec(10)-tan(10)]/[32*x].

Now substituting this and the second equation into the second identity
gives a single equation in x:

25*[16*sec(10)-tan(10)]^2/[1024*x^2] + 25/[1024*x^2] = 1,
25*[16*sec(10)-tan(10)]^2/1024 + 25/1024 = x^2,
5*sqrt([16*sec(10)-tan(10)]^2 + 25)/32 = x.

If I did my calculations correctly, the solutions are approximately

x = 2.515872
y = 3.560696 degrees.

Then angle <BAC = 86.439314 degrees.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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