Question

∆<a+<b+<c=180°,<a=60°,<b=2×c,<c=?
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Answer 1

Answer:

frnd I can't understand this question

Answer 2

Answer:

Let the <c be x

Sum of all angles in a triangle is 180deg

<a +<b +<c = 180

60+ 2×x + x =180 (becoz <b = 2×<c)

60+ 2x +x =180

3x = 180 - 60

3x = 120

$x = \frac{120}{3} \\ = > x =40$

<b = 2×x = 2×40 = 80 deg

< c = x = 40deg

Answer 3

Answer:

Let the <c be x

Sum of all angles in a triangle is 180deg

<a +<b +<c = 180

60+ 2×x + x =180 (becoz <b = 2×<c)

60+ 2x +x =180

3x = 180 - 60

3x = 120

$x = \frac{120}{3} \\ = > x =40$

<b = 2×x = 2×40 = 80 deg

< c = x = 40deg