Solving Right Triangles

Date: 03/25/98 at 08:53:00
From: Barbara Adkison
Subject: Solving right triangles

How do you figure out the cos, sin, or tan in these problems?
It says solve each triangle described. Round measures of sides to the 
nearest hundredth and measures of angles to the nearest minute.

Here is what you are given:

1.  A = 63  a = 9.7
    B = ?   b = ?
    C = 90  c = ?

How do you know when to use tan, sin, or cos?

Date: 03/25/98 at 15:59:03
From: Doctor Rob
Subject: Re: Solving right triangles

When you know two of the angles, it is easy to get the third:
A + B + C = 180.  Now draw the triangle and label the parts you know 
with their values:

               o B
              /|
             / |
            /27|
           /   |
          /    |
         /     | 9.7
        /c    a|
       /       |
      /        |
     /         |
    /63  b   90|
 A o-----------o C

Now if you know a leg and the opposite angle, you are in position to 
use a sine to find the hypotenuse and/or a tangent to find the 
adjacent leg:

   sin(A) = a/c

and you know a and A, so you can find c; and

   tan(A) = a/b

and you know a and A, so you can find b. The trick is to use an 
equation in which you know all but one part, and to use it to find the 
remaining unknown part.

If you know a leg and the adjacent angle, you are in position to use
a cosine to find the hypotenuse and/or a tangent to find the
opposite leg:

   cos(A) = b/c
   tan(A) = a/b

and you know b and A, so you can find c and a.

If you know both legs, you can use the Pythagorean theorem to find c:

   c^2 = a^2 + b^2,

and a tangent to find A:

   tan(A) = a/b.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/