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# In ∆ PQR above, w =

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Math Expert
Joined: 02 Sep 2009
Posts: 54371
In ∆ PQR above, w =  [#permalink]

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17 Aug 2017, 23:47
1
00:00

Difficulty:

15% (low)

Question Stats:

100% (00:57) correct 0% (00:00) wrong based on 41 sessions

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In ∆ PQR above, w =

(A) 50
(B) 55
(C) 60
(D) 65
(E) 75

Attachment:

2017-08-18_1028.png [ 10.41 KiB | Viewed 731 times ]

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Re: In ∆ PQR above, w =  [#permalink]

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18 Aug 2017, 01:07
Bunuel wrote:

In ∆ PQR above, w =

(A) 50
(B) 55
(C) 60
(D) 65
(E) 75

Attachment:
2017-08-18_1028.png

Small Triangle
Sum of Two unknown angles = 180-140= 40
In bigger triangle PQR
Sum of all angles = 180 = w+35 + 40 + sum of two unknown angles of smaller triangle
180 = w+35+40+40
w=65
D
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Luckisnoexcuse
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Joined: 13 Jul 2015
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Re: In ∆ PQR above, w =  [#permalink]

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18 Aug 2017, 02:12
Ans- D -65

Let the unknwon angles be x and y ,

Considering the smaller traingle , x+y+ 140 = 180.
x+y = 40

Considering the bigger triangle , 35+x+40+y+w = 180 .

Substituting hte value of x+y , w = 65.
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Re: In ∆ PQR above, w =  [#permalink]

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19 Aug 2017, 03:24
Let the Inner triangles unknown angles be x and y,

1. For inner triangle,
x+y+140=180 (sum of all angles in a triangle)

x+y=40 --(1)

2. For triangle PQR,
Sum of all angles = 35+x+y+40+w=180
using (1)
35+40+40+w=180
w=65
Re: In ∆ PQR above, w =   [#permalink] 19 Aug 2017, 03:24
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