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Byju's Answer

Standard IX

Mathematics

Converse of Theorem 5

In the given ...

Question

In the given figure, lines $\leftarrow \to A B$ and $\leftarrow \to C D$ intersect at O. If $∠ A O C + ∠ B O E = 70^{o}$ and $∠ B O D = 40^{o}$ , find $∠ B O E$ and reflex $∠ C O E$ .

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Solution

$∠ A O C + ∠ B O E = 70^{\circ}$
$40^{\circ} + ∠ B O E = 70^{\circ} ∠ B O E = 70^{\circ} - 40^{\circ} = 30^{\circ}$
$∠ C O D = 180^{\circ}$ (Angles made on line $C D$ )
reflex $∠ C O E$ $= ∠ C O D + ∠ D O B + ∠ B O E$
reflex $∠ C O E = 180^{\circ} + 40^{\circ} + 30^{\circ}$
$∠ C O E = 250^{\circ}$

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In the given figure, lines ←→AB and ←→CD intersect at O. If ∠AOC+∠BOE=70o and ∠BOD=40o, find ∠BOE and reflex ∠COE.

∠AOC+∠BOE=70∘40∘+∠BOE=70∘∠BOE=70∘−40∘=30∘∠COD=180∘ (Angles made on line CD)reflex ∠COE =∠COD+∠DOB+∠BOEreflex ∠COE=180∘+40∘+30∘∠COE=250∘

In the given figure, lines $\leftarrow \to A B$ and $\leftarrow \to C D$ intersect at O. If $∠ A O C + ∠ B O E = 70^{o}$ and $∠ B O D = 40^{o}$ , find $∠ B O E$ and reflex $∠ C O E$ .