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Question Number 71533 by pete last updated on 16/Oct/19

Three interior angles of a nonagon are  equal and the sum of the other six is 1050°  Find the size of one of the equal angles.

$$\mathrm{Three}\:\mathrm{interior}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{nonagon}\:\mathrm{are} \\ $$$$\mathrm{equal}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{six}\:\mathrm{is}\:\mathrm{1050}° \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equal}\:\mathrm{angles}. \\ $$

Answered by MJS last updated on 17/Oct/19

the sum of the interior angles is 1260°  ((1260°−1050°)/3)=70°

$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{angles}\:\mathrm{is}\:\mathrm{1260}° \\ $$$$\frac{\mathrm{1260}°−\mathrm{1050}°}{\mathrm{3}}=\mathrm{70}° \\ $$

Commented by pete last updated on 18/Oct/19

Thanks very much sir

$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

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