i need trig help

**tangerine_tomato** · June 14th 2008, 01:17 PM

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

**sean.1986** · June 14th 2008, 01:24 PM

5)

c = sqrt(53)

CosA = 2/sqrt(53)

6)

tanA = 7/6

**Reckoner** · June 14th 2008, 02:09 PM

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 $a$ and $b$ , but we'll need to use the Pythagorean theorem to get the hypotenuse:

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

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

**tangerine_tomato** · June 14th 2008, 06:29 PM

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 $a$ and $b$ , but we'll need to use the Pythagorean theorem to get the hypotenuse:

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

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

oooh o k i just i didnt understand the directions

**tangerine_tomato** · June 14th 2008, 06:30 PM

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 $a$ and $b$ , but we'll need to use the Pythagorean theorem to get the hypotenuse:

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

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

how did you draw the triangle

how did you know where the b and the a went

**Reckoner** · June 14th 2008, 06:48 PM

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 $a$ goes opposite angle $A$ , side $b$ opposite angle $B$ , etc., and the hypotenuse is usually $c$ .

**sean.1986** · June 14th 2008, 07:49 PM

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.

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