Question

in the finger o is the Central of the circle m angle aob = 75° then find
(1)m(arc A X B)
(2)m(ARC A Y B)​

Answer 1

Answer:

$\red{ m( \arc {AYB} )}\green {= 37.5\degree }$

$\red{ m( \arc {AXB} )}\green {= 142.50\degree }$

Step-by-step explanation:

$Given \: \angle {AOB} = 75\degree$

$ii) m( arc\: AYB ) = \frac{\angle{AOB}}{2}$

$= \frac{75}{2} \\= 37.5\degree$

$i) m( \arc AXB ) = 180\degree - 37.5\degree \\= 142.50\degree$

Therefore.,

$\red{ m( \arc {AYB} )}\green {= 37.5\degree }$

$\red{ m( \arc {AXB} )}\green {= 142.50\degree }$

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Answer 2

Answer:

Answer:

\red{ m( \arc {AYB} )}\green {= 37.5\degree }m(\arcAYB)=37.5°

\red{ m( \arc {AXB} )}\green {= 142.50\degree }m(\arcAXB)=142.50°

Step-by-step explanation:

Given \: \angle {AOB} = 75\degreeGiven∠AOB=75°

ii) m( arc\: AYB ) = \frac{\angle{AOB}}{2}ii)m(arcAYB)=

2

∠AOB

\begin{gathered}= \frac{75}{2} \\= 37.5\degree \end{gathered}

=

2

75

=37.5°

\begin{gathered} i) m( \arc AXB ) = 180\degree - 37.5\degree \\= 142.50\degree \end{gathered}

i)m(\arcAXB)=180°−37.5°

=142.50°

Therefore.,

\red{ m( \arc {AYB} )}\green {= 37.5\degree }m(\arcAYB)=37.5°

\red{ m( \arc {AXB} )}\green {= 142.50\degree }m(\arcAXB)=142.50°

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