Question

Help please, solve for 'x' ! explain how to do this please?

Answer 1

Hint: Use the ratios of sine, cosine, and tangent!

Answer 2

If we know the adjacent side of the given angle, and we want to find the hypotenuse, which ratio can we use?

Answer 3

Uh, im not sure..

Answer 4

Which ratio involves the adjacent and hypotenuse?

Answer 5

\[\cos 27=x/34\]

Answer 6

@mustry, why would you do that without letting her understand?

Answer 7

Anyways, the cosine ratio involves the adjacent and hypotenuse sides of the triangle.

Answer 8

we should find \[\cos \theta=hypotenuse/adjacent\]

Answer 9

So, if we set up an equation, we get something like @mustry's.

Answer 10

So if we solve for x:
\[x=34\cos27\]

Answer 11

\[\theta=27rightarrowhyp=x rightarrowadj=34\]

Answer 12

Do you understand now, @DaVaine?

Answer 13

\[x=34\cos27\]

Answer 14

The fact of the matter is that in trig you have to remember:
SOH CAH TOA

1) SOH: \(\sin(\text{angle}) = \text{opposite}/\text{hypotenuse}\)

2) CAH: \(\cos(\text{angle}) = \text{adjacent}/\text{hypotenuse}\)

3) TOA: \(\tan(\text{angle}) = \text{opposite}/\text{adjacent}\)

Answer 15

Then it is just a matter of identifying what the information given to you is.

The angle is \(27^\circ \).
The adjacent side is \(24\).
The hypotenuse is \(x\).

Since we have the adjacent side and hypotenuse, it becomes clear we want to use the CAH formula: \(\cos(27^\circ ) = 24 / x\)

Then it's just a matter of using the algebra you where taught.

Answer 16

@DaVaine Does this make any sense to you?