Answer:
Step-by-step explanation:
Here,[tex]< AOB= < AOC+ < COB[/tex][tex]< AOB[/tex]= [tex]90^{0}[/tex]
⇒[tex]< AOC+ < COB=90^{0} \\[/tex]
OR, [tex](2y-5)+(y-10)=90^{0}[/tex]
or, [tex]2y-5+y-10=90^{0}[/tex]
or, [tex]3y-15=90^{0}[/tex]
or, [tex]3y=105^{0}[/tex]
or, y=[tex]35^{0[/tex]Answer: [tex]35^{0}[/tex]
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Answers & Comments
Answer:
Step-by-step explanation:
Here,
[tex]< AOB= < AOC+ < COB[/tex]
[tex]< AOB[/tex]= [tex]90^{0}[/tex]
⇒[tex]< AOC+ < COB=90^{0} \\[/tex]
OR, [tex](2y-5)+(y-10)=90^{0}[/tex]
or, [tex]2y-5+y-10=90^{0}[/tex]
or, [tex]3y-15=90^{0}[/tex]
or, [tex]3y=105^{0}[/tex]
or, y=[tex]35^{0[/tex]
Answer: [tex]35^{0}[/tex]