Trigonometry (1 Viewer)

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FDownes

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I'm stuck on this trigonometry question, can anyone help me out? It asks;

In the figure, ABCD is a rhombus in which AB = 60cm, and angle A = 60o. P is a point on AD and AP = AQ = 2cm. Angle PQC = xo. P is joined to C.

a) Write down the length of PQ, gibing reasons.

b) Show that PC = QC

c) Use the cosine rule for the triangle QBC and, using part b), show that cos xo = √(76)/76

 
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tacogym27101990

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FDownes said:
I'm stuck on this trigonometry question, can anyone help me out? It asks;

In the figure, ABCD is a rhombus in which AB = 60cm, and angle A = 60o. P is a point on AD and AP = AQ = 2cm. Angle PQC = xo. P is joined to C.

a) Write down the length of PQ, gibing reasons.

b) Show that PC = QC

c) Use the cosine rule for the triangle QBC and, using part b), show that cos xo = √(76)/76

Click to expand...
a) PQ = 2cm triangleAPQ is equilateral

b)actually i think rather , construct CE such that angle CEQ = 90* (deg)
and i thinkkk that because Tri APQ is equilateral CE will bisect PQ and you can prove congruency in tri CEP and CEQ

c)then use cos rule to find CQ remember QB=58, BC=60
then use right angle tri CEQ to find an expression for cosx*

EDIT: the bisecting thing is correct for part b) as you can then incorporate it into c) like it asks
 
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tommykins

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回复: Re: Trigonometry

a) done
b) since APQ is equilateral, prove <CQP = <CPQ which then makes triangle CPQ isosceles -> base angle same -> sides equal.
c) had a few ideas, but it's messy, will get back to it.
 
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how can cos x be √(76)/76? look how small the adjacent is compared to the hypotenuse!!
 
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lolokay said:
how can cos x be √(76)/76? look how small the adjacent is compared to the hypotenuse!!
Click to expand...
The adjacent does look small on my diagram. Not the ratio, but when I cut the isosceles triangle into 2, it becomes 1.


ARGH I got using cosine rule that QC=√10444 o.0 Ugly surd much? Ugly surd usually equals wrong answer because I got cosx=1/(2√2611).

I did QB=58, BC=60, angle QBC=120. Used cosine rule to find QC. Divided triangle PQC so cosx=1/√10444=1/(2√2611). Can't be simplified right?


Well whatever I did wrong, I think for the answer you just need to find QC, and cosx=1/QC.
 
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tacogym27101990 said:
a) PQ = 2cm triangleAPQ is equilateral
b)prove congruency in triangles PDC and QBC should be SAS
actually i think rather , construct CE such that angle CEQ = 90* (deg)
and i thinkkk that because Tri APQ is equilateral CE will bisect PQ and you can prove congruency in tri CEP and CEQ
c)then use cos rule to find CQ remember QB=58, BC=60
then use right angle tri CEQ to find an expression for cosx*

not 100% sure about the bisecting part tho
Click to expand...
Bisecting should work because PCQ is isosceles as proven in b. I couldn't get the answer though o.0 is cosine rule for side QC^2=58^2+60^2-2x58x60xcos120?
 
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ahhliss, your working is correct (same as mine at least) so I'm pretty sure there is something wrong with the question - maybe the values are incorrect?
 
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FDownes

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Maybe, which would explain why I was having such a hard time solving this. The solutions for the papers I'm working on have proven rather unreliable.

I'll ask my maths teacher next time I get a chance.
 
bored of sc

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Using the cosine rule:

Let cos x = cos A

cos A = b2 + c2 - a2 / 2bc

PQ = 2 (from part (a))

PC = QC (from part (b))

align the sides to letters a, b and c

a = PC, b = PQ, c = QC (b and c are interchangable)

therefore:
cos A = (PQ)2 + (QC)2 - (PC)2 / 2 (PQ) (QC)

since PC = QC the equation can simplified to

cos A = (PQ)2 / 2 (PQ) (QC)

cos A = PQ / 2QC

Sub in PQ = 2

cos A = 2 / 2QC

cos A = 1 / QC

If that is even correct up to there.

So now I reckon QC is root 76 somehow :p.
 
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hmm a really good summary :D

I have a few questions about trig stuff

Prove the following identities:

cos2x-sinxcosx+tanx = 1+tan3x
cos2x+sinxcosx-tanx 1-tan3

btw the bottom two are not linked to each other its just iono how to put one under the other...

1+cscx+cotx = cscx+cotx
1+cscx-cotx
 
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Aplus

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Fortian09 said:
cos2x-sinxcosx+tanx = 1+tan3x
cos2x+sinxcosx-tanx 1-tan3
Click to expand...
Let t = tan ϑ
cos2x-sinxcosx+tanx = 1+tan3x
cos2x+sinxcosx-tanx 1-tan3

LHS = {[(1-t2)/(1+t2)]2 - [2t/(1+t2)][(1-t2)/(1+t2)] + [2t(1-t2)]} / {[(1-t2)/(1+t2)]2 + [2t/(1+t2)][(1-t2)/(1+t2)] - [2t(1-t2)]}

RHS = (1+[2t/1-t2) / (1- [2t/1-t2)

Show by expansion and simplification that LHS = RHS (cbf doing)
 
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tommykins

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回复: Re: Trigonometry

Aplus said:
Let t = tan ϑ
cos2x-sinxcosx+tanx = 1+tan3x
cos2x+sinxcosx-tanx 1-tan3

LHS = {[(1-t2)/(1+t2)]2 - [2t/(1+t2)][(1-t2)/(1+t2)] + [2t(1-t2)]} / {[(1-t2)/(1+t2)]2 + [2t/(1+t2)][(1-t2)/(1+t2)] - [2t(1-t2)]}

RHS = (1+[2t/1-t2) / (1- [2t/1-t2)

Show by expansion and simplification that LHS = RHS (cbf doing)
Click to expand...
t = tan(x/2), not tan(x)
 
tommykins

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Fortian09 said:
hmm a really good summary :D

I have a few questions about trig stuff

Prove the following identities:

cos2x-sinxcosx+tanx = 1+tan3x
cos2x+sinxcosx-tanx 1-tan3

btw the bottom two are not linked to each other its just iono how to put one under the other...

1+cscx+cotx = cscx+cotx
1+cscx-cotx
Click to expand...
Bleah anything to avoid doing English -

Cancels out to make (1+tan³x)/(1-tan³x)

I'll do the next one if you're still struggling, but think about the box a little bit.
 
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Fortian09 said:
hmm a really good summary :D

I have a few questions about trig stuff

Prove the following identities:

cos2x-sinxcosx+tanx = 1+tan3x
cos2x+sinxcosx-tanx 1-tan3

btw the bottom two are not linked to each other its just iono how to put one under the other...

1+cscx+cotx = cscx+cotx
1+cscx-cotx
Click to expand...
first one

c[c^2 - sc + t]/c[c^2 + sc - t]
= [c^3 - s[1 - s^2] + s]/[c^3 + s[1 - s^2] - s]
= [c^3 + s^3]/[c^3 - s^3]
= [1 + tan3]/[1 - tan3]

second
[s/s + 1/s + c/s]/[s/s + 1/s - c/s]
cancel out s and multiply top and bottom by [s + 1 - c]
= [s^2 + c^2 +1 + 2(sc + s c)]/[s^2 + 2s + 1 - (1-s^2)]
= 2[sc + s + c + 1]/[2s^2 + 2s]
= 2(s+1)(c+1)/2s[s+1]
= c/s + 1/s
= cotx + cscx
 
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hehe tommykins talk about messy handwriting :p
but thansk for the thelp
 
tommykins

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Started getting annoyed that I didn't specify sinx = s and cosx = c. :)
 
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FDownes

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Time for another question;

A triangle PQR is right-angled at Q. S is the point on PR such that QS is perpendicular to PR. Let angle RPQ = x. You are given that 2PS - QR = PR.

a) Show that 2cosx - tanx = secx.

b) Deduce that 2sin2 + sinx - 1 = 0.

c) Find x.
 
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