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In the figure above, DA＝DB＝DC, what is the value of x?

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In the figure above, DA＝DB＝DC, what is the value of x? [#permalink]

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In the figure above, DA＝DB＝DC, what is the value of x?

(A) 10
(B) 20
(C) 30
(D) 40
(E) 50


do you approach this problem using property of isosceles triangle?

[Reveal] Spoiler:
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[Reveal] Spoiler: OA

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Originally posted by fozzzy on 23 Jan 2013, 02:45.
Last edited by Bunuel on 28 Aug 2015, 08:50, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In the figure above, DA＝DB＝DC, what is the value of x? [#permalink]

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Don't know about what others think but I shall approach this problem using the property of isosceles triangles.
The two equal sides will have the same measure in angles too.
Since DA=DC, hence angle DAC must be 50. Similarly, angle ABD must be 30. Thus, angle BDA=180-60 or 120 and angle ADC=180-100=80.
Angle BDC is 120+80=200.
Also, since DB=DC, hence angle DBC= angle DCB and angle BDC=360-200=160.
160+2x=180, therefore x=10.
Hope that helps.
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Re: In the figure above, DA＝DB＝DC, what is the value of x? [#permalink]

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fozzzy wrote:
In the figure above DA=DB=DC. What is the value of x?

a)10
b)20
c)30
d)40
e)50


do you approach this problem using property of isosceles triangle?


Yes. Property of isosceles triangle and the sum of the internal angles of a triangle.

50 + 50 + 30 + 30 + x + x = 180

x = 10
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Re: In the figure above, DA＝DB＝DC, what is the value of x? [#permalink]

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New post 09 Jan 2018, 04:24
As DB, DA and DC are equal and all originating from D, so we can draw a circle with centre D and all three DB, DA and DC can act as radius.
Now as per circle rule, angle BAC will be half of angle BDC. So directly we can say angle BDC as 2x80 = 160 and now x = (180-160)/2 = 10.

Is this approach fine????

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Re: In the figure above, DA＝DB＝DC, what is the value of x? [#permalink]

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Isosceles triangle property and the sum of the internal angles of a triangle=180.
50 + 50 + 30 + 30 + x + x = 180
x = 10
Re: In the figure above, DA＝DB＝DC, what is the value of x?   [#permalink] 10 Jan 2018, 01:42
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