Question - 4 : - Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.

Answer - 4 : -

Let each of the two equal angles = x

Then third angle = x + 30°

But sum of the three angles of a triangle is 180°

∴ x + x + x + 30° = 180°

⇒ 3x + 30° = 180°

⇒3x = 150° ⇒x =

= 50°

∴ Each equal angle = 50°

and third angle = 50° + 30° = 80°

∴ Angles are 50°, 50° and 80°

Question - 5 : - If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.

Answer - 5 : -

In the triangle ABC,

∠B = ∠A + ∠C

But ∠A + ∠B + ∠C = 180°

⇒∠B + ∠A + ∠C = 180°

⇒∠B + ∠B = 180°

⇒2∠B = 180°

∴ ∠B =

= 90°

∵ One angle of the triangle is 90°

∴ ∆ABC is a right triangle.

Question - 6 : -
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.

Answer - 6 : -

(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180° and if there are two right-angles, then the third angle will be zero which is not possible.

(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180° and if there are

two obtuse angle, then the third angle will be negative which is not possible.

(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.

(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.

(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.

(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.

Question - 7 : - The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10°, find the three angles.

Answer - 7 : -

Let three angles of a triangle be x°, (x + 10)°, (x + 20)°

But sum of three angles of a triangle is 180°

∴ x + (x+ 10)° + (x + 20) = 180°

⇒ x + x+10°+ x + 20 = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° – 30° = 150°

∴ x =

= 50°

∴ Angle are 50°, 50 + 10, 50 + 20

i.e. 50°, 60°, 70°

Question - 8 : - ABC is a triangle is which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.

Answer - 8 : -

In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O

Now ∠B + ∠C = 180° – 12° = 108°

∵ OB and OC are the bisectors of ∠B and ∠C respectively

∴ ∠OBC + ∠OCB =

(B + C)

x 108° = 54°

But in ∆OBC,

∴ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 54° + ∠BOC = 180°

∠BOC = 180°-54°= 126°

According to corollary,

∠BOC = 90°+

∠A

= 90+

x 72° = 90° + 36° = 126°

Question - 9 : - The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Answer - 9 : -

In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠C respectively

To prove : ∠BOC cannot be a right angle

Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively

∴ ∠BOC = 90° x

∠A

Let ∠BOC = 90°, then

∠A = O

⇒∠A = O

Which is not possible because the points A, B and C will be on the same line Hence, ∠BOC cannot be a right angle.

Question - 10 : - If the bisectors of the base angles of a triangle enclose an angle of 135°. Prove that the triangle is a right triangle.

Answer - 10 : -

Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°

To prove : ∆ABC is a right angled triangle

Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O

∴ ∠BOC = 90°+

∠A

But ∠BOC =135°

∴ 90°+

∠A = 135°

⇒

∠A= 135° -90° = 45°

∴ ∠A = 45° x 2 = 90°

∴ ∆ABC is a right angled triangle

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