Question

In the adjoining figure, ∠ABD = 132° and ∠EAC = 120°. Prove that AB > AC.

Answer 1

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

Answer 2

Answer:

this question from which book tell me