# Math Help - Solving triangle ABC using given measurements

1. ## 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?

2. 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?

3. yeah so A=50 B=40 C=90
a is not the hypotenuse, c is

4. 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

5. 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

6. lengths - a = 14cm
- b = 16.68cm
-c = 21.78cm

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

7. 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?

8. 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