Q12 of 69 Page 10

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Given,

Each angle of a triangle is less than the sum of the other two.


Therefore,


∠A + ∠B + ∠C


∠A + ∠A < ∠A + ∠B + ∠C


2∠A < 180o [Sum of all angles of a triangle]


∠A = 90o


Similarly,


∠B < 90o and ∠C < 90o


Hence, the triangle is acute angled.


More from this chapter

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10

In a Δ ABC, ∠ABC =∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Shoe that ∠A =∠B =∠C = 60°.

 11

Can a triangle have:

(i) Two right angles?


(ii) Two obtuse angles?


(iii) Two acute angles?


(iv) All angles more than 60°?


(v) All angles less than 60°?


(vi) All angles equal to 60°?


Justify your answer in each case.

 1

The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

 2

In a ΔABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.