3d trig (1 Viewer)

[naq69]

AT TWO POINTS A AND B, 400 METRES APART, ON A STRAIGHT HORIZONTAL ROAD, he summit of a hill is observed. At A it is due N with an elevation of 40 degrees; at B it is due W with an elevation of 27 degrees. Find the height of the hill to the nearest 0.1 metre.

 

[ohexploitable]

Haven't done trig in AGES!!!

I got 174.2m.

 

[ohexploitable]

Here's how I did it:

Let base of hill = P
Let height of hill = h

Find PA in terms of h
Find PB in terms of h

Since there's a right-triangle, PA²+PB²=400².
Substitute the values of PA and PB in terms of h.
Make h the product.
Stick it in your calcumalator.

 

[iRuler]

Just a heads up, 3D trig = Worst part of maths

 

[twistedrebel]

there is one problem, if A and B are on horizontal line, how can the summit of the hill be due W, but do N at the same time. thats impossible or i m reading the question wrong. Source of question?

 

[ohexploitable]

there is one problem, if A and B are on horizontal line, how can the summit of the hill be due W, but do N at the same time. thats impossible or i m reading the question wrong. Source of question?

Umm... you're in flat world... in 3D world, horizontal doesn't mean constant y value, it means constant z value.

 

[twistedrebel]

Umm... you're in flat world... in 3D world, horizontal doesn't mean constant y value, it means constant z value.

have you attempted to draw it? if you have post it here so i can see

 

[kcqn93]

just a heads up, 3d trig = worst part of maths

+1

 

[lyounamu]

I got h = 55.08932... = 55.09 to 2 d.p.

Draw the diagram first. I will explain roughly how mine looks like. You first have the right angle triangle with points called A, B and O (which is the foot of the hill). Then u have two right triangles that come out of BO and AB. (caution: it's 3D diagram).

then my working out goes:

tan27 = h/BO
BO = h/tan27

and tan40 = h/AO
AO = h/tan40

but BO^2 + AO^2 = 400^2 (pythagoras' theorem)
so h^2(1/tan^2(40) + 1/tan^2(27)) = 160000
h=174.2

 

[naq69]

ohexploitable got it right
but i still dont get how the sketching would look like

 

[twistedrebel]

ohexploitable got it right
but i still dont get how the sketching would look like

What text is this?

 

[naq69]

What text is this?

not too sure
the teacher photocopied out of some text book as cambridge dont have 3d trigonometry

it says
(problems on Heights and Distances - 3 Dimensions) on top and thats all
and the exercise is
EXERCISE SET 1E

 

[ohexploitable]

ohexploitable got it right
but i still dont get how the sketching would look like

 

[lyounamu]

OMG LOL. I said that it is OA^2 + OB^2 = 16000 instead of 160000 HAHA. no wonder

 

[naq69]

very nice
thanks for the sketchin+provin

 

[hscishard]

When do we learn 3d trig? Do you really just have to know how to draw the North south east and west 3d-ishly to just be able to do it?

Because its not in the fitzpatrick book (can't find it) or Cambridge (can't find it)

 

[cutemouse]

Just a heads up, 3D trig = Worst part of maths

No it's not... It's one of the easiest topics.

They're pretty much all the same.

 

[Lolsmith]

It's generally the same as normal trigonometry, you just have to interpret the question (draw a diagram) and do a bit more working out.

 

[twistedrebel]

Just a heads up, 3D trig = Worst part of maths

there all pretty much the same

When do we learn 3d trig? Do you really just have to know how to draw the North south east and west 3d-ishly to just be able to do it?

Because its not in the fitzpatrick book (can't find it) or Cambridge (can't find it)

fitzpatrick: 21 (a) but the HSC ones are shitloads easier. HSC always give the diagram just do the working out

 

[iRuler]

there all pretty much the same

No it's not... It's one of the easiest topics.

They're pretty much all the same.

Maybe its just me, i hate 3D trig