In the adjoining figure, ∠ABD = 132° and ∠EAC = 120°. Prove that AB > AC.
Answers
Relation of Angles and Sides in a triangle.
Answer: Proved that AB > AC
Explanation:
Given that there are two exterior angles ∠DBA = 132 ° and ∠EAC= 120°.
Need to prove that AB > AC.
Angle opposite to side AB is ∠ACB and angle opposite to side AC is ∠ABC.
so first we will calculate two angles ∠BCA and ∠ABC.
using theorem which says that "In a triangle , Exterior angle is equal to sum of its two opposite interior angles , we can say that
∠BAC + ∠BCA = 132° -----eq(1) and
∠ABC + ∠BCA = 120° -----eq(2)
Adding eq(1) and eq(2) , we gwt
∠BAC + ∠BCA +∠ABC + ∠BCA = 132° + 120°
=>( ∠BAC + ∠BCA+ ∠ABC ) + ∠BCA = 252°
As ∠BAC , ∠BCA , ∠ABC are three interior angles of traingle ABC , so using angle sum property of triangle we can say that ∠BAC +∠BCA+∠ABC = 180°.
=> 180° + ∠BCA = 252°
=> ∠BCA = 252 - 180
=> ∠BCA = 72°
Now calculate ∠ABC by substituting ∠BCA = 72° in eq(2).
=> ∠ABC + 72° = 120°
=> ∠ABC = 120° - 72°
=> ∠ABC = 48°
As ∠BCA = 72° and ∠ABC = 48° , and 72° > 48°
=> ∠BCA > ∠ABC
There is one Important propety of traingle which says that "Side opposite to greater angle is longer that side opposite to smaller angle"
so as ∠BCA > ∠ABC
=> side opposite to ∠BCA > side opposite to ∠ABC
=> AB > AC
And Here Comes the most satisfying word when you solve a geometric condition that is "HENCE PROVED".
#answerwithquality
#BAL
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