Trigonometry (1 Viewer)

[Fortian09]

cheers for the advice mate
hope u go well in ur hsc

 

[FDownes]

Okay, here's another quetion;

Solve the equation cos 2x = sin x for 0 < x < 2pi.

I do have a worked solution to this question, but I don't fully understand it, so could you please explain your working in detail?

 

[tommykins]

回复: Re: Trigonometry

cos^2 x - sin^2 x = sinx
cos^2 x - sin^2 x - sin x = 0
(1-sin^2 x) - sin^2x - sin x = 0
1 - 2sin^2 x - sin x = 0

This forms a quadratic, let m = sin x and do it from there.

 

[kaz1]

Re: 回复: Re: Trigonometry

cos^2 x - sin^2 x = sinx
cos^2 x - sin^2 x - sin x = 0
(1-sin^2 x) - sin^2x - sin x = 0
1 - 2sin^2 x - sin x = 0

This forms a quadratic, let m = sin^2 x and do it from there.

Letting m=sin x would make it a whole lot easier.

 

[tommykins]

回复: Re: 回复: Re: Trigonometry

ah, silly mistake.

 

[FDownes]

Thanks. For some reason, the working in the book was completely different to the method you used, I still don't quite understand it...

 

[Fortian09]

I have a sine rule qn...
not sure how to do it...

A tower AB is 35m high. The angles of elevation of the top D of a hill from the foot A and from the top B of the tower are 46deg and 27deg respectively. Find:

a) The distance AD,
b) The horizontal distance between D and the tower AB.
Give your answer correct to nearest m.

Cheers

 

[lyounamu]

I have a sine rule qn...
not sure how to do it...

A tower AB is 35m high. The angles of elevation of the top D of a hill from the foot A and from the top B of the tower are 46deg and 27deg respectively. Find:

a) The distance AD,
b) The horizontal distance between D and the tower AB.
Give your answer correct to nearest m.

Cheers

Draw a diagram first:

where you have AB and DO (where O is the bottom). D is the apex.

DO must be taller than AB. And the angles of elevation from A to D should be 46 and angle of elevation form B to D should be 27 degrees.

To be honest, I reckon finding AO (horizontal distance from AB to DO)is the easiest thing.

tan 46 = DO/AO
DO = AO . tan 46

tan 27 = DH/AO (where H is the point on DO where the B is connected to)
DH = AO . tan 27

But DO - DH = 35
SO AO . tan46 - AO . tan 27 = AO(tan46 - tan 27) = 35
AO = 66.5393086...
Therefore, horizontal distance is 66.54 to 2 d.p.

Then you can find AD via:

cos 46 = AO/AD
AD = AO/cos46 = 95.7870968...

 

[Fortian09]

hmm ok

I have a problem with the wording in this question

A ship sails due east at 20km/h. at certain point A, the bearing of a lighthouse B from her is 045^o. after an hour, the bearing of B from the position of the ship C is 015^o. find the distance BC correct to 3sig.figs.

 

[cwag]

View attachment 17172 heres your diagram for the above question. remember bearings are measured from north in a clockwise direction

 

[Fortian09]

So basically, the ship is originally at A and after an hour of sailing, the ship is now at C is that correct?

 

[cwag]

well yea.....point A is the first place where they give you information. So for the means of this calculation..that is sufficient

 

[Fortian09]

cool thanks



Gosh the wordings of these questions are annoying, i have no idea how to do this qn

A building AB ius on the top of a terrace. from the point D on the level ground, the angles of elecation of the foot A and the top B are found to be 48^o and 65^o respectively. when measured at another point E 100m further away from the terrace, the angle of elevation of B is 35^o. Find

a) the height of the building,
b) the height of the terrace

I duno how to do this qn either

At a certain moment, a ship P at a spot A observes another ship Q at a spot B in the direction N60^oE. ship Q sails at 30km/h due north. if the speed of ship P is 30rt3km/h, what should its direction of motion be so that it can intercept ship Q?



How would you do this questions?

In triABC, /_A=30^o, /_B - 45^o and a+b+c = 3+rt2+rt3. solve triABC

 

[cwag]

View attachment 17174heres the first one...i don't know if im right...the question is a bit annoying to read....but..you can find combined height of building and terrace by 100tan35. The length of D to bas of terrace would be 100tan35/tan 65. The the height of terrance would be that answer times tan48 and hieght of building is 100tan35 - that height?? but i dunno..its seems wrogn to me

 

[Fortian09]

I actually think that the 100m is 100metres away from point D

we need namu or tommy to clarify

 

[cwag]

yea that would make sense....but the questions reads another point E 100m further away from the terrace,...i dunno...anyway...heres the second one... the speeds of the ships can be assumed to be the ratio of the sides no matter what time. View attachment 17175 you can find angle CAB by sine rule. then the bearing would be 60 - that

 

[Fortian09]

any idea how to find the @?
How would you find the angle CAB using sine rule, we need at least one angle right?

 

[cwag]

angle CBA = 120 (cointerior angles) therefore sin CAB/30 = sin 120 /30rt3

 

[Fortian09]

oooo smart thinking i didnt think of that

nice got it

 

[cwag]

i don't know how orthodox this method is...haha...but for your last one..

a + b + c = 3 + rt3 + rt2...(1)
and a/sin 30 = b/sin 45 = c/sin 105
thus, 2a=rt2b=c/sin 105...(2)

so from (2), b = rt2a ..(3)
and c = sin 105. 2a...(4)

therefore subbing into (1)
a(1 + rt2 + 2sin105) = 3 + rt2 + rt5
a= (3 + rt2 + rt5)/(1 + rt2 + 2sin105)
this miraculously comes to rt2 (see told u it was unorthodox)
therefore b = 2 (by subbing a into (3)
and c = 1 + rt3 (bu subbing a and b into (1)