Q. 3D Trig (1 Viewer)

[Preditor.89.N7G]

Ok i'm kinda new to this after doing 2unit last yr and doing a bridging course so here it goes.

Q1) ABCD is a triangular pyramid with BC = 5m, CD =8m, BC = 11m, AB =AC and /_ACB = 78 Degrees
a) Find /_BCD *
b) length AB (to nearest metre)*

Q2)An aeroplane flying at 2000m is observed to be on a bearing of 314 Degress T with an angle of elevation of 29 degrees. After 1 min, it's bearing 020 Degrees T at an angle of elevation of 25 degrees Find

a) Distance AE (already done)
b) Distance DE (done)
c) /_AED*
d) Hence find the distance the plane travels in that minute and its speed in km/h*

the 1s with * means im having trouble with them. Also with 3D trig, if any1 could add any key pointers in doing them effiecently as opposed to the normal 2unit 2D Trig.

Thankd

 

[airie]

Uhh, I think you'd need to fix your questions...

Q1) ABCD is a triangular pyramid with BC = 5m, CD =8m, BC = 11m, AB =AC and /_ACB = 78 Degrees

Huh? What exactly is BC equal to?

Also, since there are only four vertices...this "triangular pyramid" is a tetrahedron, right?

If I could take that you meant BC=5m, and BD=11m, then just use cosine rule on triangle BCD to get angle BCD. And since AB=AC and angle ACB is 78 degrees, angle ABC is the same, thus angle BAC is 180-2*78=24 degrees. Then just use sine rule on triangle ABC to find AB.

Q2)An aeroplane flying at 2000m is observed to be on a bearing of 314 Degress T with an angle of elevation of 29 degrees. After 1 min, it's bearing 020 Degrees T at an angle of elevation of 25 degrees Find

a) Distance AE (already done)
b) Distance DE (done)
c) /_AED*
d) Hence find the distance the plane travels in that minute and its speed in km/h*

Where exactly are the points A, D, and E? You never mentioned them in the question...

Basically, when doing 3D trig, just try to reduce it to 2D trig problems. Like in the first question, look for what angles/sides you need to find the quantities wanted, and just focus on those. For example in a), (assuming my guess for the lengths was correct ) you don't need the facts that AB=AC and angle ACB=78 degrees to find angle BCD, so don't be distracted by them.

Hope that helped

 

[ianc]

You've mistyped question 1 - there are 2 measurements for BC.

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In Question 2, you first start by working out the bearings. I am assuming that "T" is a typo for "E". I am also assuming that A is the position of the aeroplane at the beginning and D is the position after a minute.

314 degrees clockwise is equal to 46 degrees anticlockwise, and the other one is simply 20 degrees.

This will be the base of the 3D figure.

Next, try and draw a diagram using the information - A is the aeroplane at the beginning, 2000m above the first bearing and D is the aeroplane after a minute 2000m above the second bearing.

X is directly below A and Y is directly below D.

Now, because you have 2 right angled triangles - AXE and DYE - you can work out distances AE, XE, DE and YE.

see if you can do the rest -

• for part c you'll use the sin rule to find angles EXY and XYE (which are equal to EAD and ADE respectively), then use angle sum of triangle to get angle AED
  
• for part d you'll use the cos rule to find XY, which is equal to AD. Then to find the speed, divide this distance by the time (1 minute), making sure you've got all your units in km and hours before dividing.

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for 3D trig problems in general, the biggest hurdle is the diagram. Once you've got that, it's a breeze - the trig is usually fairly simple, at most involving the sin and cos rules.

 

[Preditor.89.N7G]

shit sorry guys, typing too fast....BC = 5M and BD =11M.
T is suppose to stand for true bearing. And thanks both ianc and airie for your tips.

 

[AlvinCY]

Q1) ABCD is a triangular pyramid with BC = 5m, CD =8m, BD = 11m, AB =AC and /_ACB = 78 Degrees


a) Find /_BCD

b) length AB (to nearest metre)


a) This question requires basic cosine rule, which states that:
cos C = (a² + b² - c²) / (2ab), so by substitution and using ∆BCD
cos (/_BCD) = (5² + 8² - 11²) / (2*5*8) = -0.4


Therefore /_BCD = cos^-1(-0.4) = 113 degrees and 34 minutes (to the nearest minute)


<!--[if !supportEmptyParas]-->b) This question requires basic sine rule, which states that:


<!--[if !supportEmptyParas]-->sin A / a = sin B / b = sin C / b where in ∆ABC, A, B, C are the three angles and a, b, c are the sides opposite the angles A, B and C respectively. (note, this is equivalent to saying a / sin A = b / sin B = c / sin C)



<!--[if !supportEmptyParas]-->So looking at ∆ABC, we see that AB=BC therefore /_ABC = 78 degrees. So /_BAC = 180 – 2*78 = 24 degrees.


Using sine rule, AB / sin 78 = 5 / sin 24, using a calculator… AB = 12m (to the nearest metre)



Q2. It can't be done unless you label where A,D and E are... like previously said, we don't know which is which. I can only make an educational guess, but even then, there are simply too many choice.



Tips for 3D Trig:
a) Ask yourself "What do I need, is it a line? Is it an angle?"
b) Look at the triangle that contains that line/angle...
c) Use 2D trig, making sure you redraw the diagram, it'll help you significantly.


Often it is easier, if you mark down every piece of information you know on the triangle involved.



Let me know what Q2's points are and I'll post you a solution