Question

The figure shows a weather balloon at point P.

A right triangle PQR is drawn with right angle at angle PRQ. Angle PQR is 35 degrees and QR is 6 miles in length.
How high is the point P of the weather balloon from a point R on the ground?

Answer 1

@Hero

Answer 2

@amistre64

Answer 3

@ash2326

@Zarkon

@thomaster

Answer 4

please help

Answer 5

@ash2326

Answer 6

You can begin by starting with what you know. We can definitely begin with

\[\tan \theta = \frac{\text{opp}}{\text{adj}}\]

Since the legs of the triangle, sides RQ and RP are the sides of interest.

Answer 7

right

Answer 8

The given angle is 35 degrees
the given adjacent side is 6
the opposite side is unknown, or \(x\)

So that yields:

\[\tan(35^{\circ}) = \frac{x}{6}\]

Solving for \(x\) yields

\[6\tan(35^{\circ}) = x\]

Do you know what \(6\tan(35^{\circ})\) is equivalent to?

Answer 9

i got 4.20124522926

Answer 10

Right, but we want the equivalent answer choice.

Answer 11

Remember, \(\tan(\theta) = \dfrac{1}{\cot(\theta)}\)

Answer 12

so the answer is c?

Answer 13

@Hero c?

Answer 14

@Hero

Answer 15

@Hero @Hero @Hero

Answer 16

It's not c, sorry. Since you have your calculator, another option for you would be to calculate each of the given options and see which one gets you same value as 4.20124522926

Answer 17

b?

Answer 18

@Hero @Hero @Hero

Answer 19

Yes, correct.

Answer 20

yay! thank you

Answer 21

The key thing to remember is
\[\tan(\theta) = \dfrac{1}{\cot(\theta)}\]

Answer 22

could you please help me with another question? @Hero