Chapter 9: Triangle and its Angles Exercise – 9.1
Question: 1
In a ΔABC, if ∠A = 55°, ∠B = 40°, Find ∠C.
Solution:
Given Data:
∠B = 55°, ∠B = 40°, then ∠C = ?
We know that
In a ΔABC sum of all angles of a triangle is 180°
i.e., ∠A + ∠B + ∠C = 180°
⇒ 55° + 40° + ∠C = 180°
⇒ 95° +∠C = 180°
⇒ ∠C = 180° − 95°
⇒ ∠C = 850°
Question: 2
If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
Solution:
Given that,
Angles of a triangle are in the ratio 1: 2: 3
Let the angles be x, 2x, 3x
∴ We know that,
Sum of all angles of triangles is 180°
x + 2x + 3x = 180°
⇒ 6x = 180°
⇒ x = 180°/6
⇒ x = 30°
Since x = 30°
2x = 2(30)° = 60°
3x = 3(30)° = 90°
Therefore, angles are 30°, 60°, 90°
Question: 3
The angles of a triangle are (x − 40°),(x − 20°) and (12x − 10°). Find the value of x.
Solution:
Given that,
The angles of a triangle are
(x − 40°),(x − 20°) and (1/2x − 10°)
We know that,
Sum of all angles of triangle is 180°
∴ (x − 40°) + (x − 20°) + (1/2x − 10°) = 180°
2x + 1/2x − 70° = 180°
5/2 x = 180° + 70°
5x = 2(250)°
x = 500°/5
∴ x = 100°
Question: 4
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°, find the three angles.
Solution:
Given that,
The difference between two consecutive angles is 10°
Let x, x+10°, x+20° be the consecutive angles that differ by 10°
We know that,
Sum of all angles in a triangle is 180°
x + x + 10° + x + 20° = 180°
3x + 30° = 180°
⇒ 3x = 180° - 30°
⇒ 3x = 150°
⇒ x = 50°
Therefore, the required angles are
x = 50°
x + 10° = 50° + 10° = 60°
x + 20° = 50° + 20° = 70°
As the difference between two consecutive angles is 10°, the three angles are 50°, 60°, 70°.
Question: 5
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.
Solution:
Given that,
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°.
Let x, x, x + 30° be the angles of a triangle
We know that,
Sum of all angles in a triangle is 180°
x + x + x + 30° = 180°
3x + 30° = 180°
3x = 180° − 30°
3x = 150°
x = 50°
Therefore, the three angles are 50°, 50°, 80°.
Question: 6
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle.
Solution:
If one angle of a triangle is equal to the sum of the other two angles
⇒ ∠B = ∠A + ∠C
In ΔABC,
Sum of all angles of a triangle is 180°
⇒ ∠A + ∠B + ∠C = 180°
⇒ ∠B + ∠B = 180°[∠ B = ∠ A + ∠C]
⇒ 2∠B = 180°
⇒ ∠B = 180°/2
⇒ ∠B = 90°
Therefore, ABC is a right angled triangle.
Question: 7
ABC is a triangle in which ∠A = 720°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.
Solution:
Given,
ABC is a triangle where ∠A = 72° and the internal bisector of angles B and C meeting O.
In ΔABC,
∠A + ∠B + ∠C = 180°
⇒ 72° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° − 72°
Dividing both sides by '2'
⇒ ∠B/2 + ∠C/2 = 108°/2
⇒ ∠OBC + ∠OCB = 54°
Now, In ΔBOC ⇒ ∠OBC + ∠OCB + ∠BOC = 180°
⇒ 540° + ∠BOC = 180°
⇒ ∠BOC = 180° − 54°=126°
∴ ∠BOC = 126°
Question: 8
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Solution:
In ΔXYZ,
Sum of all angles of a triangle is 180°
i.e., ∠X + ∠Y + ∠Z = 180°
Dividing both sides by '2'
Now in ΔYOZ
∴ ∠YOZ + ∠OYZ + ∠OZY = 180°
Therefore, the bisectors of a base angle cannot enclosure right angle.
Question: 9
If the bisectors of the base angles of a triangle enclose an angle of 135°, prove that the triangle is a right angle.
Solution:
Given the bisectors of the base angles of a triangle enclose an angle of 135°
i.e., ∠BOC = 135°
But, We know that
⇒ ∠A = 45°(2)
⇒ ∠A = 90°
Therefore, ΔABC is a right angle triangle that is right angled at A.
Question: 10
In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Solution:
Given,
In ΔABC,
∠ABC = ∠ACB
Dividing both sides by '2'
⇒ ∠OBC = ∠OCB [ ∴ OB, OC bisects ∠B and ∠C]
Now,
⇒ 30°∗ (2) = ∠A
⇒ ∠A = 60°
Now in ΔABC
∠A + ∠ABC + ∠ACB = 180° (Sum of all angles of a triangle)
⇒ 60° + 2∠ABC = 180° [∴ ∠ABC = ∠ACB]
⇒ 2∠ABC = 180° − 60°
⇒ ∠ABC = 120°/2 = 60°
⇒ ∠ABC = ∠ACB
∴ ∠ACB = 60°
Hence Proved.
Question: 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.
Solution:
(i) No, Two right angles would up to 180°. So the third angle becomes zero. This is not possible, so a triangle cannot have two right angles. [Since sum of angles in a triangle is 180°]
(ii) No, A triangle can't have 2 obtuse angles. Obtuse angle means more than 90° So that the sum of the two sides will exceed 180° which is not possible. As the sum of all three angles of a triangle is 180°.
(iii) Yes, A triangle can have 2 acute angles. Acute angle means less the 90° angle.
(iv) No, Having angles more than 60° make that sum more than 180°. This is not possible. [Since the sum of all the internal angles of a triangle is 180°]
(v) No, Having all angles less than 60° will make that sum less than 180° which is not possible.[Therefore, the sum of all the internal angles of a triangle is 180°]
(vi) Yes, A triangle can have three angles equal to 60°. Then the sum of three angles equal to the 180°. Such triangles are called as equilateral triangle. [Since, the sum of all the internal angles of a triangle is180°]
Question: 12
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
Given each angle of a triangle less than the sum of the other two
∴ ∠X + ∠Y + ∠Z
⇒ ∠X + ∠X < ∠X + ∠Y + ∠Z
⇒ 2∠X < 180° [Sum of all the angles of a triangle]
⇒ ∠X < 90°
Similarly ∠Y < 90° and ∠Z < 90°
Hence, the triangles are acute angled.