Trigonometry

**Flay** · August 1st 2008, 02:58 AM

Can anyone help me out with this question? It's really the third part that I'm having trouble with, but I'll type out the whole thing here. It asks;

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

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

b) Show that PC = QC

c) Yse the cosine rule for the triangle QBC and, using part b), show that $x^o = \frac{\sqrt{76}}{76}$ .

**Tom1234** · August 1st 2008, 03:31 AM

Originally Posted by Flay

Can anyone help me out with this question? It's really the third part that I'm having trouble with, but I'll type out the whole thing here. It asks;

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

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

b) Show that PC = QC

c) Yse the cosine rule for the triangle QBC and, using part b), show that $x^o = \frac{\sqrt{76}}{76}$ .

so
we know that triangle APQ is an equilateral triangle because it has a 60 degrees angle.
a)which means that PQ=AP=AQ=2cm.
b)you can use congreunce of triangles PDC and CBQ. where you have 2 equal lines DP and QB
BC=DC
and 2 equal opposite angles in the rhombus.
then you got b)PC=QC
c) looking at triangle QBC:
QB=60-2=58
BC=60
angle B=120
then you get QC with the cosine rule.
then you use the cosine rule in triangle PQC

**Flay** · August 1st 2008, 03:23 PM

Thanks, but my problem is that whenever I try and figure out the length of QC, I get $\sqrt{10444}$ instead of $\sqrt{76}$ . Can someone please give me a worked solution to this?

**Soroban** · August 1st 2008, 05:07 PM

Hello, Flay!

I'm getting the same answer . . .

And are you sure of the exact wording of the problem?

That last statement doesn't make sense: . $x^o \:=\:\frac{\sqrt{76}}{76}$

First of all, $\sqrt{76}$ can be simplified: . $2\sqrt{19}$

Second, that can't be the answer: . $x \:=\:0.1147^o$ ?
. . We can see that $x$ is nearly a right angle.

They may have meant: . $\cos x \:=\:\frac{\sqrt{76}}{76}$
. . But that is not the correct answer either.

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

As Tom pointed out, $APQ$ is an equilateral triangle: . $PQ = 2$

We find that: . $\angle D = 120^o,\;PD = 58,\;DC = 60$

Using the Law of Cosines on $\Delta PDC$ , we have:

. . $PC^2 \;=\;58^2 + 60^2 - 2(58)(60)\cos120^o \;=\;3364 + 3600 - 6960\left(\text{-}\frac{1}{2}\right) \;=\;10,444$

Hence: . $PC \;=\;\sqrt{10,444} \;=\;2\sqrt{2611}$

Use the Law of Cosines on $\Delta PQC$

. . $\cos x \;=\;\frac{PQ^2 + QC^2 - PC^2}{2(PQ)(QC)} \;=\;\frac{2^2 + (2\sqrt{2611})^2 - (2\sqrt{2611})^2}{2(2)(2\sqrt{2611})}$

. . $\cos x \;=\;\frac{1}{2\sqrt{2611}} \;=\;0.009785129$

Therefore: . $x \;=\;89.4393444 \;\approx\;89.44^o$

**Flay** · August 1st 2008, 06:51 PM

Yeah, that's what I was getting. I think the question paper must be wrong, because I certainly haven't written down the problem incorrectly.

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