1. Using Tangent: Word Problem

Hello everyone, today I am faced with a confusing trigonometry world problem. I need help in building a visual image of the world problem and it is been very hard for me to do that. I know the answer is 21.3m I need to know how to get there. All right here its:

From the top of a cliff 32m high it is noted that the angles of depression of two boats lying in the line due east of the cliff are 21degrees and 17degrees. How far are the boats apart?

Thank You For the Help!

2. Attached is the required image:

3. Draw a diagram with 2 triangles .

tan21 =32/x

tan 17=32/y

y-x = ans

Draw a diagram with 2 triangles .

tan21 =32/x

tan 17=32/y

y-x = ans
Alternative method:

$tan 69^o=\frac{OB_1}{32}$
$tan 73^o=\frac{OB_2}{32}$

Distance= $OB_2-OB_1$

5. Originally Posted by alexmahone
Incorrect.

$tan 69^o=\frac{OB_1}{32}$
$tan 73^o=\frac{OB_2}{32}$

Distance= $OB_2-OB_1$

6. Originally Posted by flutterby
Refer attachment in the second post.

Note that the complements of the angles 21degrees and 17degrees are 69degrees and 73degrees respectively.

7. so is it like you switch to the angle of elevation and then solve from there?

8. Originally Posted by flutterby
so is it like you switch to the angle of elevation and then solve from there?
Not exactly.

Refer this link: Angles of Elevation and Depression

9. Originally Posted by alexmahone
Not exactly.

Refer this link: Angles of Elevation and Depression

mmm can you please explain how it is more like?

10. Originally Posted by alexmahone
Incorrect.

$tan 69^o=\frac{OB_1}{32}$
$tan 73^o=\frac{OB_2}{32}$

Distance= $OB_2-OB_1$
Hi Alex .

Angle PCB2=angle CB2O ( from the diagram you posted )

so tan 21 = 32/x and same goes to the other one .

Look at example 21 in the link , i did what exactly it says .

Hi Alex .

Angle PCB2=angle CB2O ( from the diagram you posted )

so tan 21 = 32/x and same goes to the other one .

Look at example 21 in the link , i did what exactly it says .
Your method works as well. Fair enough. Sorry.

Hi Alex .

Angle PCB2=angle CB2O ( from the diagram you posted )
Do you mean $\angle OCB_2 = \angle CB_2O$? If so, this is incorrect. $\angle OCB_2$ is not the angle that is labeled as 17°. However, $\angle CB_2O = 17^{\circ}$ is correct.

Anyway, it looks like both you and alexmahone are correct in your working out this problem. It also looks like the y that mathaddict assigns is the same as the side $OB_2$.

According to alexmahone:
$\tan 73^{\circ} = \frac{OB_2}{32}$.

$\tan 17^{\circ} = \frac{32}{OB_2}$.

But,
$\tan (90^{\circ} - \theta) = \cot \theta$

So,
\begin{aligned}
\tan 73^{\circ} &= \tan (90^{\circ} - 17^{\circ}) \\
\frac{OB_2}{32} &= \cot 17^{\circ} \\
\frac{32}{OB_2} &= \tan 17^{\circ}
\end{aligned}

... meaning that both of you are saying the same thing.

01

EDIT: Beaten to it