Question

What is the height, in meters, of the river dam?

145 tan 65°

145 cos 65°

145 by tan 65 degrees

145 by cos 65 degrees

Answer 1

@mathstudent55 can you please help me with this?

Answer 2

Look at the 65-deg angle.
For that angle, the height is the adjacent or the opposite leg?

Answer 3

The adjacent? And thank you by the way for helping @mathstudent55

Answer 4

You're welcome, and you're correct.
Since the height is the adjacent leg, that means the 145-m side is the opposite leg since it is certainly not the hypotenuse, right?

Answer 5

Yes. (just to clarify, is "m" just the unknown number or does it signify something?)

Answer 6

m means meters. That leg of the triangle is 145 meters long.

Answer 7

oh ok:)

Answer 8

We are looking for the height, which is the left vertical leg of the triangle.

Answer 9

Ok, we have a right triangle, a known acute angle (65 deg), a known opposite leg (145 m) and an unknown adjacent leg (height or h).

Answer 10

Of the three main trigonometric functions, sine, cosine, and tangent, which one relates the adjacent and opposite legs? (Hint: do you know SOHCAHTOA?)

Answer 11

Umm...is it cosine?

Answer 12

No.
Let's review the basic trig functions.
I referred to SOHCAHTOA above.
Did you ever hear of that?

Answer 13

kind of

Answer 14

Ok, we'll review that, just to make sure you go from "kind of" to "I know it."

Answer 15

isn't CAH stand for cosine equals adjacent over Hypotenuse.

Answer 16

*doesn't

Answer 17

The trig functions are ratios of lengths of sides of a right triangle.

SOHCAHTOA reminds you of the ratios with the initials.
You are correct about the cosine.
Here are the three of them.

\(\sin \theta = \dfrac{opp}{hyp}\) SOH

\(\cos \theta = \dfrac{adj}{hyp}\) CAH

\(\tan \theta = \dfrac{opp}{adj}\) TOA

Answer 18

So then it must be tangent, correct?

Answer 19

Now in our problem we have the known opposite and the unknown adjacent.
We need a function that has opp and adj in it.
Which one is it?

Answer 20

Correct.
We us the tangent.

Answer 21

\(\tan 65^o = \dfrac{145~m}{h} \)

Answer 22

So do we further simplify that equation or no?

Answer 23

Yes, we have to simplify it. In fact, we need to solve for h, the height, because that is what the problem is asking for.

Now we solve for h.
We can write tan 65 as a fraction and cross multiply.

\(\dfrac{\tan 65^o}{1} = \dfrac{145~m}{h} \)

\(h \tan 65^o = 145~m\)

Divide both sides by tan 65 to get:

\(h = \dfrac{145~m}{ \tan 65^o }\)

Answer 24

So you finish off with a cross multiplication. Thank you so much for helping! I understand it now!!!

Answer 25

You're welcome.