# i need trig help

• Jun 14th 2008, 01:17 PM
tangerine_tomato
i need trig help
Two sides of a right triangle ABC (C is the right angle) are given. Find the indicated trigonometric function of the given angle.

5) Find cosA when a=7 an b=2

6) Find tanA when a=7 and b=6

thank you :)
• Jun 14th 2008, 01:24 PM
sean.1986
5)

c = sqrt(53)

CosA = 2/sqrt(53)

6)

tanA = 7/6
• Jun 14th 2008, 02:09 PM
Reckoner
Quote:

Originally Posted by tangerine_tomato
5) Find cosA when a=7 an b=2

6) Find tanA when a=7 and b=6

Hello again, tangerine. We usually label the sides of a right triangle with the lowercase letters corresponding to their opposite angles (as in my drawing).

So,

5. We know $\displaystyle a$ and $\displaystyle b$, but we'll need to use the Pythagorean theorem to get the hypotenuse:

$\displaystyle \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac bc = \frac2{\sqrt{7^2 + 2^2}} = \frac2{\sqrt{53}} = \frac{2\sqrt{53}}{53}$

6. $\displaystyle \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac ab = \frac76$
• Jun 14th 2008, 06:29 PM
tangerine_tomato
Quote:

Originally Posted by Reckoner
Hello again, tangerine. We usually label the sides of a right triangle with the lowercase letters corresponding to their opposite angles (as in my drawing).

So,

5. We know $\displaystyle a$ and $\displaystyle b$, but we'll need to use the Pythagorean theorem to get the hypotenuse:

$\displaystyle \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac bc = \frac2{\sqrt{7^2 + 2^2}} = \frac2{\sqrt{53}} = \frac{2\sqrt{53}}{53}$

6. $\displaystyle \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac ab = \frac76$

oooh o k i just i didnt understand the directions
• Jun 14th 2008, 06:30 PM
tangerine_tomato
Quote:

Originally Posted by Reckoner
Hello again, tangerine. We usually label the sides of a right triangle with the lowercase letters corresponding to their opposite angles (as in my drawing).

So,

5. We know $\displaystyle a$ and $\displaystyle b$, but we'll need to use the Pythagorean theorem to get the hypotenuse:

$\displaystyle \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac bc = \frac2{\sqrt{7^2 + 2^2}} = \frac2{\sqrt{53}} = \frac{2\sqrt{53}}{53}$

6. $\displaystyle \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac ab = \frac76$

how did you draw the triangle

how did you know where the b and the a went
• Jun 14th 2008, 06:48 PM
Reckoner
Quote:

Originally Posted by tangerine_tomato
how did you draw the triangle

how did you know where the b and the a went

Really, I didn't know for sure since the problem didn't specify. But as I said before, that is the usual convention: side $\displaystyle a$ goes opposite angle $\displaystyle A$, side $\displaystyle b$ opposite angle $\displaystyle B$, etc., and the hypotenuse is usually $\displaystyle c$.
• Jun 14th 2008, 07:49 PM
sean.1986
The right angle is always opposite the longest side (hypoteneuse).

So in this case, the hypoteneuse would be c, right angle would be C.

The other two angles A and B are oppsite sides a and b respectively. The positioning of these two sides is irrelevant because you're basically just flipping or turning the same triangle.