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Question
In the given figure, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
Options
15°
25°
70°
35°
Solution
In the given figure,
PQ || RS
Also,∠AEFand ∠1 are the corresponding angles.
Then, according to the Corresponding Angles Axiom, which states:
If a transversal intersects two parallel lines, then each pair of corresponding angles are equal.
Therefore,
∠1 = ∠AEF
It is given that
∠AEF = 95°
Therefore,
∠1 = 95°
Clearly, ∠1 and ∠2form a linear pair, therefore, their sum must be supplementary.
Therefore,
∠1 + ∠2 = 180°
On substituting ∠1 = 95°in equation above, we get:
95° + ∠2 =180°
∠2 = 180° - 95°
∠2 = 85°
In ΔBHG:
We know that, in a triangle exterior angle is equal to the sum of the interior opposite angles. Therefore,
∠2+x = ∠BHS
Substituting ∠BHS =110°
and ∠2 = 85°, we get :
85° + x = 110°
x = 110° - 85°
x = 25°
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