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arhumsid

What is the perimeter of ∆PQS ?

(1) x = 45
(2) w = 15


Because angle w is outside of ∆PQS, it is tempting to think that angle w is not required to construct ∆PQS,
but it is. The second paragraph below shows that w can determine x. Hence, if Statement (1) alone can
answer the question, then Statement (2) can as well.
The minimum details needed to construct a triangle are one side and two angles, or two sides and one
angle, or all the three sides. In ∆PQS, we have ∠PSQ = 30° (from the figure) and the side PS = 1. So, we
are short of knowing one angle or one side to construct ∆PQS. Statement (1) would help with an angle in
the triangle. So, Statement (1) is sufficient.
Construct line PS = 1. Draw an angle making 30 degrees with PS at S and name the line l. Extend the line
PS 2 units further to the right of the point S to locate the point R. Now, draw another line measuring w (=
15) degrees with PR from the point R, and name the line m. So, the point of intersection of l and m is the
point Q. Now, measure ∠PQS to find x.
The answer is (D).


I will try,

The Question is whether we can draw a fixed Triangle ∆PQS or not.

As we know to get a fixed Triangle, we need atleast two sides and one included angle or two angles and one included side.

Statement 1): We are given ∠x=45 then ∠p=180-45-30=105.
Now we have Two angles and one included side so we have fixed triangle.
we will have fixed perimeter for ∆PQS.

Hence , Sufficient.

Statement 2): We are given ∠w=15 and also we know ∠y=150, so ∠(q-x)=15=∠w.
now ∆SQR is Isosceles trainlge, if side opposite to ∠(q-x) ,SR, is 2 mm then side opposite to ∠w, QS, is also 2 mm.
Now in ∆PQS, we have two sides and one included angle , so we have one fixed trainlge ∆PQS.
we will have fixed perimeter for ∆PQS.

Hence, Suffiecient.

And) D

please correct me if i am wrong.
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arhumsid

What is the perimeter of ∆PQS ?

(1) x = 45
(2) w = 15


Because angle w is outside of ∆PQS, it is tempting to think that angle w is not required to construct ∆PQS,
but it is. The second paragraph below shows that w can determine x. Hence, if Statement (1) alone can
answer the question, then Statement (2) can as well.
The minimum details needed to construct a triangle are one side and two angles, or two sides and one
angle, or all the three sides. In ∆PQS, we have ∠PSQ = 30° (from the figure) and the side PS = 1. So, we
are short of knowing one angle or one side to construct ∆PQS. Statement (1) would help with an angle in
the triangle. So, Statement (1) is sufficient.
Construct line PS = 1. Draw an angle making 30 degrees with PS at S and name the line l. Extend the line
PS 2 units further to the right of the point S to locate the point R. Now, draw another line measuring w (=
15) degrees with PR from the point R, and name the line m. So, the point of intersection of l and m is the
point Q. Now, measure ∠PQS to find x.
The answer is (D).

I think Nova is a good book to practice but the questions sometimes can be a bit too calculation intensive or ask to apply very specific formulae. This is one such question.

Statement 1 is sufficient as mentioned in posts above.

For statement 2, once you figure out that QSR is an isosceles triangle giving you QS=2. You are already given PS=1 and \(\angle {QSP} = 30\).

Remember the "rule" that you can create a fixed triangle with 2 sides and the included angle.

But for the proof of it, you will have to use 3 trigonometric formulae (not recommended for GMAT)

1. \(sin^2(x) + cos^2(x) = 1\)
2.\(sin (x+y)= sin(x)*cos(y)+sin(y)*cos(x)\)
3. \(a/sin(x)=b/sin(y)=c/sin(z)\)

Rest assured you will be able to find unique value for PQ leading to a unique perimeter for triangle PQS.

Hope this helps.
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trigonometry helps a a lot:

(1) x=45

the area of a triangle equals 1/2*a*b*sinA

since you know all the three angles and one line, all the length of three lines can be obtained:

1: 1/2*PQ*1*sin105=1/2*PQ*QS*sin45

so you have QS

then 1/2*QS*1*sin30=1/2*PQ*QS*sin45

so you have PQ

sufficient

(2) y=150, w=15, so QSR is an isosceles triangle and now you have QS=2

when you have two length and its angle, you can get the third:

PQ^2=QS^2+PS^2-2QS*PS*sin30

sufficient
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Thanks for the responses! That's some quality discussion you guys have come up with.

So in conclusion, there 2 basic things we need to realize to answer this question:
1) As with every DS question, we dont have to 'find' the answer. We only need to prove that is can be found.
2) The 'sufficiency' to obtain a unique triangle-
Having two sides and 1 included angle
Having two angles and 1 included side
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Bunuel,

Could you help to solve this question.

Thanks,
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