In Fig. 10.15, find the sum of $∠ 1, ∠ 2, ∠ 3, ∠ 4, ∠ 5, ∠ 6, ∠ 7, ∠ 8 a n d ∠ 9.$

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Solution

In triangle AED, using angle sum property
$∠ 7 + ∠ 8 + ∠ 9 = 180^{\circ}$

In triangle ACD, using angle sum property
$∠ 4 + ∠ 5 + ∠ 6 = 180^{\circ}$

In triangle ABC, using angle sum property
$∠ 1 + ∠ 2 + ∠ 3 = 180^{\circ}$

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Similar questions

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

(1), (2, 3, 4), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8, 9, 10)

\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}

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