Question 1: In a triangle ΔABC, if ∠A = 55° ∠B = 40°, find ∠C.
Solution:
Given: A = 55° and B = 40°
Theorem Used: Sum of all angles of a triangle is 180°.
From the theorem we can write that:
∠A + ∠B + ∠C = 180°
55° + 40° + ∠C = 180° //Putting the values of A and B.
95° + ∠C = 180°
∠C = 180° – 95°
∠C = 85°
The angle ∠C is 85°.
Question 2: If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
Given: Angles of a triangle are in the ratio 1:2:3
Let the angles be x, 2x, 3x
Theorem Used: Sum of all angles of a triangle is 180°.
x + 2x + 3x = 180°
6x = 180°
x = 180°/6
x = 30° //Deriving the value of x
Deriving the value of the other two angles from the value of x
2x = 2 X (30°) = 60°
3x = 3 X (30°) = 90°
All the three angles are 30°,60° and 90° respectively.
Question 3: The angles of a triangle are (x − 40)°, (x − 20)° and (1/2 x − 10)°. Find the value of x.
Solution:
The angles of a triangle are (x − 40)°, (x − 20)° and (1/2x − 10)°
Theorem Used: Sum of all angles of a triangle is 180°.
(x − 40)° + (x − 20) ° + (1/2 x − 10)° = 180°
5/2 x – 70° = 180°
5/2 x = 180° + 70°
5x = 2(250)°
x = 500°/5
x = 100°
The value of x is 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: The difference between two consecutive angles is 10°.
Theorem Used: Sum of all angles of a triangle is 180°.
Let the smallest angle of the triangle is x°.
Hence, according to the given condition, the other two consecutive angles are (x + 10)° and (x + 20)° respectively.
Now from the theorem mentioned we can write that:
x + (x + 10°) + (x + 20°) = 180°.
3x + 30° = 180° //Simplifying the equation
3x = 180° – 30°
3x = 150°
x = 150°/3
x = 50°
Hence, here we get the smallest angle is 50°.
The next consecutive angles are, 50° + 10° = 60° and 50° + 20° = 70° respectively.
Hence, the three angles of the triangle are 50°, 60°, and 70° respectively.
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: (i) Two angles of a triangle are equal
(ii)The third angle is greater than each of those angles by 30°
Theorem Used: Sum of all angles of a triangle is 180°.
Let the equal angles are x° and the other angle is (x+30)°.
Now from the theorem mentioned we can write that:
x + x + (x + 30°) = 180°
3x + 30° = 180°
3x = 180° – 30°
3x = 150°
x = 150°/3
x = 50°
Hence, the equal angles are 50° each and the other angle is (50 + 30)° = 80°.
The angles of the triangle are 50°, 50° and 80° respectively.
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:
Given: one angle of a triangle is equal to the sum of the other two
Theorem Used: Sum of all angles of a triangle is 180°.
Let the three angles of the triangle are ∠A, ∠B and ∠(A+B) respectively.
∠A + ∠B + ∠(A + B) = 180°
2(∠A + ∠B) = 180°
∠A + ∠B = 90° //Hence the third angle A + B = 90° (Proved)
Question 7: ABC is a triangle in which angle ∠A = 72°. The internal bisector of angle ∠B and ∠C meet in O. Find the magnitude of ∠BOC.
Solution:
Given: (i) ∠A = 72° of the triangle ABC
(ii)Internal bisectors of angle ∠B and ∠C meet in point O.
Theorems Used: Sum of all three angles of a triangle is 180°
In triangle ABC,
∠A + ∠B + ∠C = 180°
72° + ∠B + ∠C = 180°
∠B + ∠C = 180° – 72° = 108°
∠B/2 + ∠C/2 = 108°/2 = 54° // Dividing both sides by 2
∠OBC + ∠OCB = 54° –derivation (1) // From the triangle we can clearly see this as OB and OC are the angle bisectors
Now in △BOC,
∠OBC + ∠OCB + ∠BOC = 180°
∠BOC + (∠OBC + ∠OCB) = 180°
∠BOC + 54° = 180° //Putting the value of ∠OBC + ∠OCB = 54° from derivation(1)
∠BOC = 180° – 54° = 126° (Ans)
∠BOC = 126°.
Question 8: The bisectors of the base angles of a triangle cannot enclose a right angle at any case.
Solution:
Theorems Used: Sum of all three angles of a triangle is 180°
From the triangle △ABC,
∠A + ∠B + ∠C = 180°
∠A/2 + ∠B /2 + ∠C/2 = 180°/2 = 90° //Dividing both sides by 2
∠B /2 + ∠C/2 = 90° – ∠A/2 —-(1)
From the triangle △BOC,
∠BOC + ∠OBC + ∠OCB = 180°
As OB and OC are the angle bisectors, ∠OBC = ∠B/2 and ∠OCB = ∠C/2.
∠BOC + ∠B /2+ ∠C/2 = 180° //Putting the values of ∠B /2+ ∠C/2 = 90° – ∠A/2 from the previous derivation,
∠BOC + 90° – ∠A/2 = 180°
∠BOC = 180° – 90° + ∠A/2 = 90° + ∠A/2
For any valid triangle △ABC ∠A > 0, It implies that ∠A/2 > 0,
That simply means
∠BOC not equals to 90° at any case. (proved)
Question 9: If the bisectors of base angles of a triangle enclose an angle of 135°, Prove that the triangle is a right-angle triangle.
Solution:
Given: In △BOC the ∠BOC = 135°
Theorems Used: Sum of all three angles of a triangle is 180°
From the triangle △ABC,
∠A + ∠B + ∠C = 180°
∠A/2 + ∠B /2+ ∠C/2 = 180°/2 = 90° //Dividing both sides by 2
∠B /2+ ∠C/2 = 90° – ∠A/2 —-(1)
From the triangle △BOC,
∠BOC + ∠OBC + ∠OCB = 180°
As OB and OC are the angle bisectors, ∠OBC = ∠B/2 and ∠OCB = ∠C/2.
∠BOC + ∠B/2 + ∠C/2 = 180° //Putting the values of ∠B/2 + ∠C/2 = 90° – ∠A/2 from the previous derivation,
∠BOC + 90° – ∠A/2 = 180°
∠BOC = 180° – 90° + ∠A/2 = 90° + ∠A/2
Putting the value ∠BOC = 135° from the given condition,
90° + ∠A/2 = 135°
∠A/2 = 135° – 90° = 45°
∠A = 45° X 2 = 90°
Hence, △ABC is a right angle triangle(proved)
Question 10: In a triangle △ABC, ∠ABC = ∠ACB and the bisector of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Solution:
Given: (i)∠ABC = ∠ACB
(ii)∠BOC = 120°
From the triangle △ABC,
∠ABC = ∠ACB
∠ABC/2 = ∠ACB/2
∠OBC = ∠OCB
From the triangle △ABC,
∠OBC + ∠OCB + ∠BOC = 180°
From the given condition ∠BOC = 120°, and ∠OBC = ∠OCB
We can write that,
∠OBC + ∠OBC + 120° = 180°.
2 X ∠OBC = 180° – 120° = 60°
∠ABC = 60°
As angle ∠ACB = ∠ABC,
∠ACB = 60°
∠ACB + ∠ABC + ∠BAC = 180°
60° + 60° + ∠BAC = 180°
∠BAC = 180° – 120° = 60°
Hence,
∠A = ∠B = ∠C = 60°.(Proved)
Question 11: Can a triangle have,
(i) Two right angles.
If the triangle have two right angles, sum of those angles become 90° + 90° = 180° , That implies that the size of the third angle is 180° – 180° = 0, That is not possible,
Answer: NO
(ii)Two obtuse angles
The size of an obtuse angle is greater than 90° , Hence the sum of both angles are greater than 180° , But we know sum of all three angles of a triangle is 180°. So it is not possible.
Answer: NO
(iii) Two acute angles
Having two acute angles does not violates any law as their sum is less than 180°
Answer: YES
(iv) All Angles More than 60°
Having all angles more than 60° will make the sum of all angles > 180°.But we know sum of all three angles of a triangle is 180° . So it is not possible.
Answer: NO
(v) All Angles Less than 60°
Having all angles more than 60° will make the sum of all angles < 180°. But we know sum of all three angles of a triangle is 180° . So it is not possible.
Answer: NO
(vi) All angles equal to 60°
Having all angles equal to 60° will make the sum of all angles = 180°.And we know sum of all three angles of a triangle is 180° . So it is possible.
Answer: YES
Question 12: If each angle of a triangle is less than the sum of the other two, Show that all angles of the triangle are acute angle.
Solution:
Given: Each angle is less than the sum of the other two
∠A + ∠B + ∠C = 180°
Given that ,∠A < ∠B + ∠C, So we can write,
∠A < 90°,
The same thing can be done for ∠B and ∠C.
Hence, it is proved that all the three angles are acute angle.