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In a ` Delta ABC, angle C = 3, angle B = 2 (angle A + angle B)`, then ` angle B ` = ? |
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Answer» Let ` angle A= x ^(@) and angle B = y^(@)`. Then, ` angle C = 3 angle B = ( 3y) ^(@)` Now, ` angle C = 2 (angle A + angle B )` ` rArr 3y = 2 (x + y ) rArr 2x - y = 0 " " `… (i) We know that the sum of the angles of a triangle is ` 180^(@)` ` therefore angle A + angle B + angle C = 180 ^(@) rArr x + y + 3y = 180` ` " " rArr x + 4y = 180" " `... (ii) On multiplying (i) by 4 and adding the result with (ii), we get ` 8x + x = 180 rArr x = 20`. Putting ` x = 20 ` in ( i ), we get `y = ( 2 xx 20 ) = 40 ` Thus, ` x = 20 and y = 40 ` ` therefore " " angle A = 20 ^(@), angle B = 40 ^(@) and angle C = ( 3xx 40 ^(@)) = 120 ^(@)` |
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