In this section, we shall discuss inequality in triangle.
Theorem -1 : If two sides of a triangle are unequal, the longer side has greater angle opposite to it.
Given : A ΔABC in which AC > AB.
Prove that : ∠ABC > ∠ACB
Construction : Mark a point D on AC such that AB = AD. Join BD.
1) AB = AD
1) By Construction
2) ∠ ABD = ∠ADB
2) If two sides are equal then angle opposite to them are also equal
3)∠ADB > ∠DCB
3)As ∠ADB is an exterior angle of ΔBCD and exterior angle is always greater than interior angle.
4)∠ADB > ∠ACB
4)∠ACB = ∠DCB
5) ∠ABD > ∠ABC
5) ∠ABC = ∠ABD + ∠DBC
6) ∠ABC > ∠ACB
6) From (4) and (5)
Converse of the above theorem is also true.
Theorem : 2 In a triangle the greater angle has the longer side opposite to it.
1) In ΔABC, AC = 5cm, AB = 7 cm and BC = 3Cm. Write the angles in ascending order.
2)In ΔPQR, PQ = 8cm, PR = 3 cm and PQ= 6Cm. Write the angles in ascending order.
3) In ΔABC, AC is the longest side then which angle is the largest ?
4) In ΔPQR, QR is the shortes side then which angle is the smallest ?
5) In a right triangle MNO, right angled at N, which side is the longest side?
6) n ΔPQR, ∠P =400,∠Q =800 and ∠R =600. Write the sides in ascending order.