in the given figure ABC is a triangle having side BC and CA produced to D and E respectively then AC>BC .is this true ?
pls answer this question it's very urgent for me.....
Answers
no it is not true
angleACD+angleACB=180
AngleACB=40
AngleBac=80
So Angle ABC=60
Angle Abc<Angle Bac
So,Ac<BC
Therefore the statement AC > BC is true for the given triangle ABC.
Given:
In the given figure, ΔABC is a triangle having side CA and BC produced to D and E respectively
such that ∠DCA = 140° and ∠BAE = 100°
To Find:
To confirm whether AC is greater than BC or not.
Solution:
The given question can be solved as shown below.
Given that,
∠DCA = 140° and ∠BAE = 100°
We know that on a straight line the total angle is always 180°.
On the line segment BCD,
⇒ ∠DCA + ∠ACB = 180°
⇒ 140° + ∠ACB = 180°
⇒ ∠ACB = 40°
On the line segment CAE,
⇒ ∠EAB + ∠BAC = 180°
⇒ 100° + ∠BAC = 180°
⇒ ∠BAC = 80°
In any triangle, the sum of the interior sides of that triangle is always 180°.
⇒ ∠BCA + ∠CAB + ∠ABC = 180°
⇒ 40° + 80° + ∠ABC = 180°
⇒ ∠ABC = 180° - 120° = 60°
So, ∠A = 80°, ∠B = 60°, ∠C = 40°
As ∠A > ∠B > ∠C ⇒ BC > AC > AB
So AC > BC
Therefore the statement AC > BC is true for the given triangle ABC.
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