Question

In the figure 10.9, / || m | n and p is a transversal. If 21 = 110°, find angles x, y, z,
and w.
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Answer 1

Answer:

x=70

y=70

z=70

w=110

Step-by-step explanation:

angle 1 and x are linear pair so if we add both of them we will 180

1+x=180

x=180-1

x=180-110=70

and angle x and angle y are exterior alternative angles

so they are equal

and angle y and angle z are interior alternative angles

so they are equal

z and w are again linear pair so they will add up-to 180

so z+w=180

70+w=180

w=180-70=110

w=110

Answer 2

➝ANSWER

●Given:— 1) ABCD is a quadrilateral.

2) $\sf\angle1 = 110°$

●To find:— $\sf\angle x \: \angle y \: \angle z \: \angle w$

●Solution:—

$\sf\angle 1 + \angle x \: = 180 ^{\circ} \: (linear \: pair) \\ \sf 110 ^{\circ}+ \angle x = 180 ^{\circ} \\ \sf \angle x =180 ^{ \circ} - 110 ^{ \circ} \\ \sf \angle x = 70 ^{ \circ}$

$\rm \: l \parallel m \: and \: p \: is \: a \: transversal \\ \sf \angle x = \angle y = 70 ^{ \circ} (alternate \: exterior \: opp. \: angle) \\ \\ \rm \: m \parallel n \: and \: p \: is \: a \: transversal \\ \sf \angle y = \angle z = 70 ^{ \circ} (alternate \: interior \: opp. \: angle) \\ \\ \rm \: l \parallel \: m \: and \: p \: is \: a \: transversal \\ \sf \angle1 = \angle w = 110 ^{ \circ} (alternate \: exterior \: opp. \: angle)$

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