Find $$<x+<y$$
here is the problem.......
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- Kazi_Zareer
- Posts:86
- Joined:Thu Aug 20, 2015 7:11 pm
- Location:Malibagh,Dhaka-1217
Re: Find $$<x+<y$$
( $360^{\circ}$ $-$ $x$ ) $+$ ( $360^{\circ}$ $ - $ $y$ ) = $720^{\circ}$ $ -$ ( $ 30^{\circ}$ $+$ $45^{\circ}$ $+$ $50^{\circ}$ $+$ $25^{\circ}$ )
$ \Rightarrow $ $x$ $+$ $y$ = $150$
$ \Rightarrow $ $x$ $+$ $y$ = $150$
We cannot solve our problems with the same thinking we used when we create them.
Re: Find $$<x+<y$$
how did you get that $720^{\circ}$ $ -$ ( $ 30^{\circ}$ $+$ $45^{\circ}$ $+$ $50^{\circ}$ $+$ $25^{\circ}$ )??Kazi_Zareer wrote:( $360^{\circ}$ $-$ $x$ ) $+$ ( $360^{\circ}$ $ - $ $y$ ) = $720^{\circ}$ $ -$ ( $ 30^{\circ}$ $+$ $45^{\circ}$ $+$ $50^{\circ}$ $+$ $25^{\circ}$ )
$ \Rightarrow $ $x$ $+$ $y$ = $150$
- Kazi_Zareer
- Posts:86
- Joined:Thu Aug 20, 2015 7:11 pm
- Location:Malibagh,Dhaka-1217
Re: Find $$<x+<y$$
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.
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Re: Find $$<x+<y$$
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˚