In the figure given L1 and L2 are parallel- The sum of the angles - - - marked in the diagram is-

The correct option is C 360^0 Extend the line AO to OA^' ⇒∠ OA^' B=180 α ⇒∠ A^' OB=180 β ⇒∠ A^' OB=180 γ As OA^' B is a triangle, ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

In the figure...

Question

In the figure given $L_{1}$ and $L_{2}$ are parallel. The sum of the angles $α, β, γ$ marked in the diagram is:

180^{0}

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270^{0}

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360^{0}

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\leq 270^{0}

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Solution

The correct option is C $360^{0}$
Extend the line AO to $O A^{^{'}}$

$\Rightarrow ∠ O A^{^{'}} B = 180 - α$

$\Rightarrow ∠ A^{^{'}} O B = 180 - β$

$\Rightarrow ∠ A^{^{'}} O B = 180 - γ$

As $O A^{^{'}} B$ is a triangle, sum of its angles =180 degrees

$\Rightarrow 540 - (α + β + γ) = 180$

$\Rightarrow α + β + γ = 360$

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In the figure given L1 and L2 are parallel. The sum of the angles α,β,γ marked in the diagram is:

The correct option is C 3600Extend the line AO to OA′⇒∠OA′B=180−α⇒∠A′OB=180−β⇒∠A′OB=180−γAs OA′B is a triangle, sum of its angles =180 degrees ⇒540−(α+β+γ)=180⇒α+β+γ=360

In the figure given $L_{1}$ and $L_{2}$ are parallel. The sum of the angles $α, β, γ$ marked in the diagram is: