Lines AB and CD interesct at O- If - AOC -135- then - BOC -3-A- 65-B- 20-C- 15-

The correct option is C 15^∘ ∠AOC + ∠BOC = 180^∘ nbsp; nbsp;∵ they form nbsp;a linear pair. nbsp;Given that, ∠AOC = 135^∘ ⇒135^∘ + ∠BOC = nbsp;180^ ...

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Byju's Answer

Standard VIII

Mathematics

Supplementary, Complementary Angles and Linear Pair Axiom

Lines AB and ...

Question

Lines AB and CD interesct at O. If $∠$ AOC = $135^{\circ}$ , then $\frac{∠ B O C}{3}$ = .

65^{\circ}

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20^{\circ}

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15^{\circ}

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Solution

The correct option is C $15^{\circ}$

$∠$ AOC + $∠$ BOC = $180^{\circ}$ $∵$ they form a linear pair.

Given that, $∠$ AOC = $135^{\circ}$

$\Rightarrow$ $135^{\circ}$ + $∠$ BOC = $180^{\circ}$

$\Rightarrow$ $∠$ BOC = $180^{\circ}$ - $135^{\circ}$

$\Rightarrow$ $∠$ BOC = $45^{\circ}$
$\Rightarrow$ $(\frac{∠ B O C}{3})$ = $15^{\circ}$ .

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Similar questions

Q. Lines AB and CD interesct at O. If

∠

AOC =

135^{\circ}

, then

\frac{∠ B O C}{3}

= .

In the figure below, a ray OC extends from a point O on a straight line AB. If $∠$ AOC = $135^{\circ}$ . Find $(\frac{∠ B O C}{3})$ .

Q. In the figure given below, a ray OC extends from a point O on a straight line AB, If ∠AOC = 135°, then

\frac{∠ B O C}{3}

is equal to:

In the figure below, a ray OC extends from a point O on a straight line AB. If $∠ A O C = 135^{\circ}$ ,

find $(\frac{∠ B O C}{3})$ .

In the given figure, straight lines AB and CD intrsect at O. If $∠ A O C = ϕ, ∠ B O C = θ$ and $θ = 3 ϕ$ , then $ϕ = ?$

$(a) 30^{\circ}$

$(b) 40^{\circ}$

$(c) 45^{\circ}$

$(d) 60^{\circ}$

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Lines AB and CD interesct at O. If ∠AOC = 135∘, then ∠BOC3 = .

The correct option is C 15∘ ∠AOC + ∠BOC = 180∘ ∵ they form a linear pair. Given that, ∠AOC = 135∘ ⇒135∘ + ∠BOC = 180∘ ⇒∠BOC = 180∘ - 135∘ ⇒∠BOC = 45∘ ⇒(∠ BOC3) = 15∘.

Lines AB and CD interesct at O. If $∠$ AOC = $135^{\circ}$ , then $\frac{∠ B O C}{3}$ = .