Question

Answer me fast please bro

Answer 1

In Δ ABC,

∠A + ∠ B = 150° ---> ( 1 )

∠ B + ∠ C = 100 ° ---> ( 2 )

⏺️∠ B = 100 - ∠ C ---> ( i )

Subtracting equation 1 and 2 ,

∠A + ∠ B - ( ∠ B + ∠C) = 150 - 100

∠A + ∠ B - ∠ B - ∠C = 50

∠A - ∠ C = 50

⏺️∠ A = 50 + ∠ C --- > ( ii )
Sum of all three ∠s of a Δ is 180 °.

∠ A + ∠ B + ∠C = 180 °

( 50 + ∠ C ) + ( 100 - ∠ C ) + ∠ C = 180°

150 + ∠ C = 180

∠ C = 180 - 150 = 30 °

Putting value of ∠ C in ( i ),

∠B = 100 - ∠ C

∠B = 100 - 30 = 70°

Putting value of ∠ C in ( ii ),

∠A = 50 + ∠ C

∠ A = 50 + 30 = 80 °

✔️ ANSWER =

∠ A = 80 °, ∠ B = 70 °, ∠ C = 30°.

Answer 2

$\tiny \bf{ < A + < B = 150 \: and < B + < C = 100} \\ \\ \tiny\bf{we \: know \: the \: sum \: of \: three \: side \: of \: trianlge \: \: is \: 180} \\ \\ \bf{ < A + < B + < C = 180} \\ \\ here \: < A + < B = 150 \: then,\\ \\150 + < c = 180 \\ \\ < c = 180 - 150 = 30 \\ \\ and \: another \: is \: < B + < C = 100 \\ \\ as \: same > > \\ \\ < A + < B + < C = 180 \\ \\ here \: < B + < C = 100 \\ \\ < A +100 = 180 \\ \\ < A = 180 - 100 = 80 \\ \\ after \: that \: we \: have \: to \: find \: < b \: so \\ \\ < A + < B + < C = 180 \\ now \: we \: got \: < A = 80 \: and < C = 30 \\ \\ 80 + < b + 30 = 180 \\ \\ 110 + < B = 180 \\ \\ < B = 180 - 110 \\ \\ = > > < B = 70$