Find $$<x+<y$$

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jayon
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Find $$<x+<y$$

Unread post by jayon » Sat Dec 17, 2016 4:49 pm

here is the problem.......
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Kazi_Zareer
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Re: Find $$<x+<y$$

Unread post by Kazi_Zareer » Sun Dec 18, 2016 12:54 am

( $360^{\circ}$ $-$ $x$ ) $+$ ( $360^{\circ}$ $ - $ $y$ ) = $720^{\circ}$ $ -$ ( $ 30^{\circ}$ $+$ $45^{\circ}$ $+$ $50^{\circ}$ $+$ $25^{\circ}$ )

$ \Rightarrow $ $x$ $+$ $y$ = $150$
We cannot solve our problems with the same thinking we used when we create them.

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jayon
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Re: Find $$<x+<y$$

Unread post by jayon » Sun Dec 18, 2016 11:55 am

Kazi_Zareer wrote:( $360^{\circ}$ $-$ $x$ ) $+$ ( $360^{\circ}$ $ - $ $y$ ) = $720^{\circ}$ $ -$ ( $ 30^{\circ}$ $+$ $45^{\circ}$ $+$ $50^{\circ}$ $+$ $25^{\circ}$ )

$ \Rightarrow $ $x$ $+$ $y$ = $150$
how did you get that $720^{\circ}$ $ -$ ( $ 30^{\circ}$ $+$ $45^{\circ}$ $+$ $50^{\circ}$ $+$ $25^{\circ}$ )??

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Kazi_Zareer
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Re: Find $$<x+<y$$

Unread post by Kazi_Zareer » Sun Dec 18, 2016 1:15 pm

Use this formula: $(n-2) \times 180^{\circ} $ for measuring the sum of the internal angles of a regular $n$-gon. Now try to think about it.
We cannot solve our problems with the same thinking we used when we create them.

aritra barua
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Re: Find $$<x+<y$$

Unread post by aritra barua » Thu Jan 19, 2017 7:23 pm

X+Y=150˚,as because by extending some lines in the figure,at a point;we find Y+30˚ as the supplementary angle of X ;so,X+Y+30=180˚,X+Y=150˚ :D

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