In the figure above, point O is the center of the semicircle, and PQ

Since all Triangles are isosceles.. ∠POQ=∠QPO=70° Hence ∠QOP=40°

∠QPO+∠SOP=180 hence ∠QOS=70

Hence 70+4x+2=360°

Hence x=72° Option D.

Regards,
Dom.

Answer will be (A) 34deg.
All triangles are isosceles because of the radius. SOP=110 and QOP=40
Since the ROS will be 2deg lesser than ROP, hence 34deg

You calculated value of 'x' and not ROS.


\(\angle{ROS}\) = \(\pi\)- 2*x-2 = 180-144-2=180-146=34

It should be A) 34 degrees
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KAPLAN OFFICIAL SOLUTION:

Analyze: What shapes do we see in this mess? A semicircle and three isosceles triangles (we know they’re isosceles because each of them has two sides that are radii of the semicircle).

Because they’re isosceles, what else can we label? We know that for all three triangles, the two “top” angles are equal. So, angle OPQ is 70°, angle OQR is x°, and ORS is x + 1°.

What else does the Q-stem tell us? That PQ is parallel to OS. That means we can find corresponding angles.

Where is a corresponding angle to angle OQP found? Angle QOS. It must also be 70°.

Why might that be of interest? Because PO and QO are both transversals.

Task: Good analysis. What are we being asked for? The measure of angle ROS.

And what do we know about triangles and their internal angles? They sum to 180°. so, < ROS + 2(x +1) = 180.

Approach strategically: All right. We know a lot about triangles QOR and ROS now. What will the sum of their combined angles be? 360. Each triangle has 180°.

Give us an equation that contains all the info we have. 360 = 70 + 4x + 2.

Solve that for x. 360 = 4x + 72 → 288 = 4x → 72 = x.

And how can we use that to calculate angle ROS? 180 = 2(72 + 1) + < ROS → 180 = 146 + < ROS → 34 = < ROS.

Answer (a) is correct.

Confirm: That wasn’t so bad actually. Notice how we used triangles, circles, and even good ol’ lines and angles to get the solution. Now, what if you didn’t have time for those steps. Is there a guessing strategy available? The triangles are pretty close and the lines are parallel, so x° must be close to 70°. That would get us down to (a) or (B). Given that <QOR has angles of x° and < ROS has angles of (x + 1)°, it makes sense to go with (a).

By the way, what’s up with answers (D) and (E)? They’re traps in case we solved for x or (x + 1) instead of what we’re supposed to.
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Pl see the attached figure..
the angles in black are given and
the ones in red have been deducted since all triangkes are isosceles triangle due to the radius..
it is given that PQ is parallel to OS..
so 40 + 180-2x+180-2(x+1)=180...
or 4x=288
x=72...
therefore angle ROS= 180-2(72+1)=34
A

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Hi All,

This question is based on a series of standard Geometry formulas, but you have to stay organized to see how they all 'interact' with one another.

1) Since each triangle consists of 2 radii, each of the triangles is ISOSCELES, meaning that the two base angles are the same. Thus, the three triangles have angles of (70 and 70), (X and X) and (X+1 and X+1).
2) Lines PQ and QS are parallel, so angle PQO (70 degrees) = angle QOS (70 degrees).
3) Quadrilateral QRSO has 4 sides, so its 4 angles total 360 degrees.

The 4 angles that sum to 360 degrees are:

70 + X + (X+X+1) + (X+1) = 360

Combining like terms gives us...

4X + 72 = 360
4X = 288
X = 72

The question asks us for angle ROS. Since the two matching angles are both (X+1), those two angles are 73 and 73, leaving the third angle...

180 - 2(73) = 180 - 146 = 34

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Since we know PQ is parallel to OS,
Hence ∠QOS = 70 (Alternate angles)
Since ∠ ROS is a part of ∠ QOS, Options D and E cannot be true.

In \(\triangle\) ROQ, ∠ROQ = ∠QRO = x (Since the two sides are radius, hence this becomes an isosceles triangle)
Hence ∠QOR = 180 - 2x (Sum of angles of a triangle = 180)

Similarly, In \(\triangle\) ROS,
∠ROS = ∠RSO = x + 1
Hence ∠ROS = 180 - 2x - 2 = 178 - 2x

Now, ∠ROS + ∠QOR = ∠QOS = 70 (By alternate angles)
Hence 180 - 2x + 178 - 2x = 70
4x = 288, x = 72

∠ROS = 178 - 2x = 178 - 144 = 34
Option A

Another way...
Sum of interior angles of OPQRS =(5-2)*180=540...

All three triangles are isosceles triangle...
So angle Q+R+S= (x+1)*2+x*2+70=4x+72.....
Angle P+O= 180, since PQ is parallel to OS...

But all above angles are 540..
So Q+R+S=4x+72=ALL-(P+O)=540-180=360......
Thus x is (360-72)/4= 72...
Angle ROS= 180-2(x+1)=180-2(72+1)=34..
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OK so i could get an answer but One silly question i have is

Considering Triangles PQO , QR0 are similar and from Triangle PQO -> Angle QPO=Angle PQO =70 , cant i say Angle qro and OQR is also = 70.?

Bunuel Can you please guide?

You cannot say that \(\triangle PQO .....and..... \triangle QRO\) are similar because we do not know the angles or the lengths PQ and RQ.
Rather if they are similar, they will be congruent too. All angles are same, so angle POQ and ROQ are same and two sides containing this angle are radius, so SAS rule. BUT we do not have any info to say they are similar or congruent.
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