Question

In the following right triangle, find sin A, cos A, tan A, and sin B, cos B, tan B.

a = ?, b = x, c = 3x

I know how to go through the process if it didn't have variables. But can someone explain the process for the above please. Thanks!

Edit: Going through the process with Angle A only should be sufficient.

Answer 1

is there any pictures

Answer 2

so that we know which side is a, b, c.

Answer 3

Here it is.

Answer 4

well, the missing leg/side will be => \(\large b = \sqrt{(3x)^2 -x^2} \implies \sqrt{8x^2} \implies 2x\sqrt{2} \)

Answer 5

once you have all those 3 guys, you can just use the so-called SOH CAH TOA identities

Answer 6

Let's call the missing side 'a'. Here we apply the pythagorean theorem:\[\bf a^2 + x^2 = (3x)^2 = 9x^2 \implies a^2=8x^2 \implies a = 2\sqrt{2}x\]Now that we have everything in terms of 'x', we can now solve for it. Now to find the trigonmetric ratios, i.e sine, cosine, tangent, we use SOHCAHTOA:\[\bf \sin(A)=\frac{ Opposite }{ Hypotenuse }=\frac{2\sqrt{2}\cancel{x}}{3\cancel{x}}=\frac{ 2\sqrt{2} }{ 3 }\]Similary, find cos(A), tan(A), etc...

Answer 7

@fleep

Answer 8

Got it, I didn't simplify it further.... guess my basic algebra needs improving Thanks!