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Question
In the given figure, AB || CD || EF and GH || KL. The measure of ∠HKL is
Options
85°
135°
145°
215°
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Solution
The given figure is as follows:
Let us extend GH to meet AB at Y.
We have the following:
∠CHG and∠CHY are the supplementary angles. Therefore,
∠CHY +∠CHG =180°
∠CHY = 180° -∠CHG
= 180° - 60°
= 120°
∠CHY = 120° .... (1)
Since,AB||CD. Thus, ∠DHKand ∠YKHare the interior alternate angles.
Therefore,
∠DHK = ∠YKH
∠YKH = 25° .... (2)
Since, AB || CD. Thus, ∠CHY and ∠AYZare the corresponding angles.
Therefore,
∠CHY = ∠AYZ
From (1), we get:
∠AYZ = 120° ……(3)
Since ∠YZ || KL. Thus ∠AYZ and ∠YKL are corresponding angles,
Therefore,
∠AYZ = ∠YKL
∠YKL = 120° .... (4)
Thus, the required angle x can be calculated as:
x = ∠YKL + ∠YKH
From (3) and (4) we get:
x = 25° + 120°
x= 145°
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