# Thread: Help to find length of Side Np

1. ## Help to find length of Side Np

Hello, I have two images here, the first of which is a diagram of the question below it.

So as you can see from the question they want the length of NP.

Now In my diagram, I equated the
angle of depression of L to N, as 28.7 degrees on the right angled triangle with hypotenuse 9m.

Now from Here, I've been trying various methods to figure out the length of NP.

I first tried, Sine28.7degrees= Np/9m

So: Sine28.7 x 9 = NP = 4.32m.

I then tried adding all the angles in that triangle, to get the unknown angle.

We have 28.7+90 = 118.7 degrees

SO therefore the other angle is = 180-118.7 degrees = 61.3 degrees.

Now using this, I did, Sine61.3 = LP/9

= Sine61.3 x 9 = LP = 7.89m

Then using my two known sides 7.89m and 9m, I used pythagoras' theorem to solve for NP

92-7.892 = NP2

NP= 4.32.

However the answer in the book says NP = 7.89m.

Which is the answer that I got when I calculated LP.

So I'm quite confused about what could be going wrong here.

2. My guess, based on your observation, is the angle of depression is between the rope and the ground, not the post.

3. First, please allow me to say "Thank you!" for showing your work and reasoning so nicely!

Originally Posted by richiesmasher

...they want the length of NP. I equated the angle of depression of L to N, as 28.7 degrees on the right angled triangle with hypotenuse 9m.
Because they state that the angles are of "depression", not "elevation", I agree with your interpretation of the location of these measured angles.

Originally Posted by richiesmasher
I've been trying various methods to figure out the length of NP.

I first tried, Sine28.7degrees= N[P]/9m

So: Sine28.7 x 9 = NP = 4.32m.
I agree. However:

Let's use what we know from "real life". If the rope is longer (as LN is longer than LM), then the angle to get there must be shallower. (A steeper angle of depression would result in the rope reaching the ground faster!) So the listed information cannot be correct; the angle of depression for LM can not be larger than the angle of depression for LN. Or else the ropes' lengths must be different from what is listed.

What happens if we reverse the angle measures?

The angles of depression of M and N from L are 28.7o and 39.5o, respectively.

Then we get:

. . . . .NP/9 = sin(39.5o)

. . . . .NP = 9*sin(39.5o) = 5.7247...

This doesn't match the book, either. So maybe the angles were correct, but the ropes' lengths were swapped...?

LM = 9 and LN = 6...

This gives us:

. . . . .NP/6 = sin(28.7o)

. . . . .NP = 6*sin(28.7o) = 2.8813...

So this doesn't match the book's answer, either.

I can only guess that there is a big typo somewhere....

Originally Posted by richiesmasher
...the answer in the book says NP = 7.89m, which is the answer that I got when I calculated LP.
So maybe the book meant to ask for the length of LP? Or whoever wrote the solutions manual copied down the wrong value? I'd ask the instructor about this. You're probably not the only one who's confused!

4. Is this the same book that contains all the other misprints you've talked about (in your other threads)?

If so, you may want to consider getting a better book.

5. Originally Posted by stapel
First, please allow me to say "Thank you!" for showing your work and reasoning so nicely!

Because they state that the angles are of "depression", not "elevation", I agree with your interpretation of the location of these measured angles.

I agree. However:

Let's use what we know from "real life". If the rope is longer (as LN is longer than LM), then the angle to get there must be shallower. (A steeper angle of depression would result in the rope reaching the ground faster!) So the listed information cannot be correct; the angle of depression for LM can not be larger than the angle of depression for LN. Or else the ropes' lengths must be different from what is listed.

What happens if we reverse the angle measures?

The angles of depression of M and N from L are 28.7o and 39.5o, respectively.

Then we get:

. . . . .NP/9 = sin(39.5o)

. . . . .NP = 9*sin(39.5o) = 5.7247...

This doesn't match the book, either. So maybe the angles were correct, but the ropes' lengths were swapped...?

LM = 9 and LN = 6...

This gives us:

. . . . .NP/6 = sin(28.7o)

. . . . .NP = 6*sin(28.7o) = 2.8813...

So this doesn't match the book's answer, either.

I can only guess that there is a big typo somewhere....

So maybe the book meant to ask for the length of LP? Or whoever wrote the solutions manual copied down the wrong value? I'd ask the instructor about this. You're probably not the only one who's confused!
Originally Posted by mmm4444bot
Is this the same book that contains all the other misprints you've talked about (in your other threads)?

If so, you may want to consider getting a better book.
Hey thanks guys, that gives me some insight then, it is true that the book has misprints, I am actually almost finished with it, there are 6 practice tests in the back of the book, this is the third one.

After this I will be doing past examinations, those don't have any answers, the exam in Particular is CSEC written in the caribbean, this is an old book.

Also I do not have an instructor stapel, I'm currently self tutoring.

6. Originally Posted by richiesmasher
Now In my diagram, I equated the angle of depression of L to N, as 28.7 degrees on the right angled triangle with hypotenuse 9m.
The angle of depression is defined as the angle from the horizontal DOWN to the line referred to, which is the same as the angle of elevation from the other end. That is, angle PLN should be marked as the complement of 28.7°, not as 28.7°, and angle PNL is 28.7°. See options 1 and 2 here.

7. Originally Posted by Dr.Peterson
The angle of depression is defined as the angle from the horizontal DOWN to the line referred to, which is the same as the angle of elevation from the other end. That is, angle PLN should be marked as the complement of 28.7°, not as 28.7°, and angle PNL is 28.7°. See options 1 and 2 here.

Dr Peterson, You are absolutely correct, it isn't a misprint after all, all the time I had this nagging feeling in my head, ''Don't I have to draw a horizontal?''

All the while completely forgetting that angles of depression form alternate angles, so therefore angle PNL is 28.7, from the LP = 4.32 after using the sine function, and from there NP is = 7.89!

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