Solving triangle ABC using given measurements

• May 4th 2008, 11:25 AM
emmahlav
Solving triangle ABC using given measurements
I was not paying attention in class when this was taught and now I'm completely lost

right triangle ABC
B=40 degrees

side a=14

what do I do next?
• May 4th 2008, 11:29 AM
Jason Holm
Well if it's a right triangle, one angle is 90. You said another is 40, and all three angles of a triangle should equal 180. Is side A the hypotenuse or one of the other sides?
• May 4th 2008, 11:32 AM
emmahlav
yeah so A=50 B=40 C=90
a is not the hypotenuse, c is
• May 4th 2008, 11:48 AM
Jason Holm
Here's the steps, but I don' know offhand how to do each of them:

If you know two angles and any side:

• Use the Sum of the Angles with the two angles to find the third angle
• Use the Law of Sines and plug in the values for the two angles and the side
• Solve for the side
• Use the Law of Sines with an angle, the side opposite it, and the angle opposite the side you still don't know to find that side
• May 4th 2008, 11:49 AM
Jason Holm
The Law of Sines defines the relationship between the sine of any angle in a triangle and the side opposite it. The formula is: a/sinA = b/sinB = c/sinC
• Jun 2nd 2008, 02:22 AM
chelsealover25
lengths - a = 14cm
- b = 16.68cm
-c = 21.78cm

angles - A = 50
- B = 40
- C = 90
• Jun 2nd 2008, 07:37 AM
masters
Quote:

Originally Posted by chelsealover25
lengths - a = 14cm
- b = 16.68cm
-c = 21.78cm

angles - A = 50
- B = 40
- C = 90

Are you sure about the values for b and c?
• Jun 2nd 2008, 07:40 AM
topsquark
Quote:

Originally Posted by emmahlav
I was not paying attention in class when this was taught and now I'm completely lost

right triangle ABC
B=40 degrees

side a=14

what do I do next?

You can use simple trigonometric definitions. No need to get fancy.
$cos(40^o) = \frac{14}{c} \implies c = 18.2757$

and
$tan(40^o) = \frac{b}{14} \implies b = 11.7474$

And you can check your answers using the Pythagorean Theorem: $14^2 + b^2 = c^2$.

-Dan