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Question
first prove the triangles are similar and properly justify your answer. then solve for side length x.
Step1: Find angles in \(\triangle ABC\)
In \(\triangle ABC\), using the angle - sum property of a triangle (\(\angle A+\angle B+\angle C = 180^{\circ}\)). Given \(\angle A = 40^{\circ}\), \(\angle C=22^{\circ}\), then \(\angle B=180^{\circ}-(40^{\circ}+22^{\circ}) = 118^{\circ}\).
Step2: Prove similarity
In \(\triangle ABC\) and \(\triangle ZXY\), \(\angle A=\angle Z = 40^{\circ}\), \(\angle B=\angle X = 118^{\circ}\). By the AA (Angle - Angle) similarity criterion, \(\triangle ABC\sim\triangle ZXY\).
Step3: Set up proportion
Since \(\triangle ABC\sim\triangle ZXY\), the ratios of corresponding sides are equal. \(\frac{AB}{ZX}=\frac{AC}{ZY}\). We know \(AB = 5\), \(ZX=\frac{50}{7}\), \(AC = 7\), and \(ZY=x\). Substituting the values into the proportion: \(\frac{5}{\frac{50}{7}}=\frac{7}{x}\).
Cross - multiply: \(5x=7\times\frac{50}{7}\).
Step4: Solve for \(x\)
Simplify the right - hand side: \(5x = 50\). Divide both sides by \(5\): \(x=\frac{50}{5}=10\).
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The value of \(x\) is \(10\).