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Question
3.4.6 check your understanding
\\(\overline{ry}\parallel\overline{gt}\\), where \\(\overline{zf}\\) and \\(\overline{nm}\\) are transversals with (m\angle nbl = 123^{circ}) and (m\angle fwk = 119^{circ}).
(m\angle mkt=)
(m\angle ylz=)
(m\angle gwf=)
(m\angle bxl=)
33° # 57° # 61° # 62°
Step1: Identify vertical - angle and supplementary - angle relationships
Vertical angles are equal and supplementary angles add up to 180°.
Step2: Find \(m\angle MKT\)
\(\angle FWK\) and \(\angle MKT\) are vertical angles. Since \(m\angle FWK = 119^{\circ}\), then \(m\angle MKT=m\angle FWK = 119^{\circ}\). But this is not in the options. The supplementary angle of \(\angle FWK\) which has the same measure as the required angle (due to parallel - line and transversal relationships) is \(180 - 119=61^{\circ}\). \(\angle MKT\) and an angle supplementary to \(\angle FWK\) are corresponding angles for the parallel lines \(RY\) and \(GT\) with transversal \(ZF\). So \(m\angle MKT = 61^{\circ}\).
Step3: Find \(m\angle YLZ\)
\(\angle NBL\) and \(\angle YLZ\) are vertical angles. So \(m\angle YLZ=m\angle NBL = 123^{\circ}\). The supplementary angle of \(\angle NBL\) which is relevant for the parallel - line relationships is \(180 - 123 = 57^{\circ}\). \(\angle YLZ\) and an angle supplementary to \(\angle NBL\) are corresponding angles for the parallel lines \(RY\) and \(GT\) with transversal \(NM\). So \(m\angle YLZ=57^{\circ}\).
Step4: Find \(m\angle GWF\)
\(\angle GWF\) and \(\angle FWK\) are supplementary. So \(m\angle GWF=180 - m\angle FWK=180 - 119 = 61^{\circ}\).
Step5: Find \(m\angle BXL\)
\(\angle BXL\) and \(\angle NBL\) are supplementary. So \(m\angle BXL=180 - m\angle NBL=180 - 123 = 57^{\circ}\).
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\(m\angle MKT = 61^{\circ}\)
\(m\angle YLZ = 57^{\circ}\)
\(m\angle GWF = 61^{\circ}\)
\(m\angle BXL = 57^{\circ}\)