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

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Math Expert
Joined: 02 Sep 2009
Posts: 51035
In the figure above, point O is the center of the semicircle, and PQ  [#permalink]

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27 Dec 2015, 09:59
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Difficulty:

75% (hard)

Question Stats:

51% (02:33) correct 49% (03:02) wrong based on 63 sessions

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In the figure above, point O is the center of the semicircle, and PQ is parallel to OS. What is the measure of ∠ ROS?

A. 34 °
B. 36 °
C. 54.5 °
D. 72 °
E. 73 °

Attachment:

2015-12-27_2157.png [ 27.68 KiB | Viewed 2946 times ]

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Posts: 7095
Re: In the figure above, point O is the center of the semicircle, and PQ  [#permalink]

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27 Dec 2015, 20:11
Bunuel wrote:

In the figure above, point O is the center of the semicircle, and PQ is parallel to OS. What is the measure of ∠ ROS?

A. 34 °
B. 36 °
C. 54.5 °
D. 72 °
E. 73 °

Attachment:
The attachment 2015-12-27_2157.png is no longer available

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
Attachments

Untitled.png [ 24.87 KiB | Viewed 2491 times ]

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Re: In the figure above, point O is the center of the semicircle, and PQ  [#permalink]

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28 Dec 2015, 21:56
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

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Re: In the figure above, point O is the center of the semicircle, and PQ  [#permalink]

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28 Dec 2015, 22:26
Given: PQ || SO.

Therefore Angle PQO = Angle SOQ = 70deg.

We are asked to find the measure of Angle ROS.

From the figure Angle SOQ = Angle QOR + Angle ROS = 70. Lets call this equation 1.

Triangles QOR and ROS are isosceles (OQ=OR=OS= Radius).

Angle QOR = 180 - 2X
Angle ROS = 180 - 2(X+1).

Substituting in equation 1, we have:

180-2X + 180- 2(X+1) = 70, solving for X we have X = 72.

and hence angle ROS = 180 - 2(X+1) = 34.
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In the figure above, point O is the center of the semicircle, and PQ  [#permalink]

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29 Dec 2015, 01:32
Bunuel wrote:

In the figure above, point O is the center of the semicircle, and PQ is parallel to OS. What is the measure of ∠ ROS?

A. 34 °
B. 36 °
C. 54.5 °
D. 72 °
E. 73 °

Attachment:
2015-12-27_2157.png

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
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Re: In the figure above, point O is the center of the semicircle, and PQ  [#permalink]

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23 Aug 2018, 07:55
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Re: In the figure above, point O is the center of the semicircle, and PQ &nbs [#permalink] 23 Aug 2018, 07:55
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# In the figure above, point O is the center of the semicircle, and PQ

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