(2)
1) in ∆ abc
angle <abc + <bca < cba = 180⁰ so
b + < c + 50⁰ = 180
180- 50 = 130
130/2 = 65
so ,angle b and angle c is 65
3)in ∆abc
<ABC + <BCA + <CAB = 180⁰
so,
<CAB + <BCA+ <ABC= 180
120 + <A + <ACB= 180
< BCA + < ACB = 180 - 120= 60
60/2 = 30
SO angle b and angle c is 30⁰
(<) I use symbol instead of angel symbol
plz make me brailiest I did very hard work for your answer
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Verified answer
(2)
1) in ∆ abc
angle <abc + <bca < cba = 180⁰ so
b + < c + 50⁰ = 180
180- 50 = 130
130/2 = 65
so ,angle b and angle c is 65
3)in ∆abc
<ABC + <BCA + <CAB = 180⁰
so,
<CAB + <BCA+ <ABC= 180
120 + <A + <ACB= 180
< BCA + < ACB = 180 - 120= 60
60/2 = 30
SO angle b and angle c is 30⁰
(<) I use symbol instead of angel symbol
plz make me brailiest I did very hard work for your answer