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If l||m and m∠6 = 4x - 15 and m∠7 = x + 30, then m∠6 = 15 45
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given line m II line l and m<6 = 4x-15 and m<7 = x+30 line t intersect line m and l so from this result that m<6 = m<7 so => 4x-15 = x+30 3x = 45 x = 45/3 x = 15 so using this value of x => m<6 = 4x-15 so m<6 = 4*15 -15 so m<6 = 45 degree hope helped
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