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In ∆ PQR above, y =

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Bunuel
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In ∆ PQR above, y = [#permalink] New post   17 Aug 2017, 23:49
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In ∆ PQR above, y =

(A) 130
(B) 120
(C) 100
(D) 80
(E) 50

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2017-08-18_1028_001.png
2017-08-18_1028_001.png [ 10.58 KiB | Viewed 1516 times ]
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Re: In ∆ PQR above, y = [#permalink] New post   18 Aug 2017, 00:05


In ∆ PQR above,
4x + 80 = 180(The sum of angles in a triangle is 180)
4x = 100
x = 25

In the smaller triangle, 2x + y = 180
Substituting the value for x, we get
50 + y = 180
y = 130(Option A)
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Re: In ∆ PQR above, y = [#permalink] New post   18 Aug 2017, 02:05
Ans a = 130

For bigger triangle , 4x+80 = 180 ; x=25
For smaller triangle , 2x+y= 180 , Substituting x=25 , y= 130
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Re: In ∆ PQR above, y = [#permalink] New post   18 Aug 2017, 02:18
In the triangle PQR

Since the sum of angles in a triangle is 180

4x + 80 = 180
4x = 100
x = 25

In the triangle inscribed inside

2x + y = 180

Substituting the value for x, we get

50 + y = 180
y = 130
(Option A)
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Re: In ∆ PQR above, y = [#permalink] New post   18 Aug 2017, 03:07
80 + 2X + 2X = 180

We get X = 25

therefore = y + 2x = 180
Y = 130 taking the value of x = 25
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Re: In ∆ PQR above, y = [#permalink] New post   19 Aug 2017, 03:19
In the Outer triangle PQR,

Sum of angles = 2x + 2x + 80 =180
Therefore x =25

In the Inner triangle,
sum of angles = x + x + y = 180
Therefore y= 180-25-25
y=130
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Re: In ∆ PQR above, y = [#permalink] New post   19 Aug 2017, 22:02
Bunuel wrote:

In ∆ PQR above, y =

(A) 130
(B) 120
(C) 100
(D) 80
(E) 50

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Attachment:
2017-08-18_1028_001.png


1. X + X + X + X + 80 (angle PQR) = 180. Solve, we get x = 25.
2. Look in the small triangle : X + X + Y = 180. Plug in X value, we get 50 + Y = 180. So Y is 130.

A.
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Re: In ∆ PQR above, y = [#permalink]
19 Aug 2017, 22:02
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