问题1：在三角形ΔABC中，如果∠A= 55°∠B= 40°，则求出∠C。
解决方案：
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°.
问题2：如果三角形的角度比例为1：2：3，则确定三个角度。
解决方案：
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.
问题3：三角形的角度为(x − 40)°，(x − 20)°和(1/2 x − 10)°。求x的值。
解决方案：
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°
问题4：三角形的角度按升序排列。如果两个连续角度之间的差为10°，则找到三个角度。
解决方案：
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.
问题5：一个三角形的两个角度相等，并且第三个角度比每个角度大30°。确定三角形的所有角度。
解决方案：
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.
问题6：如果三角形的一个角度等于其他两个角度的和，则表明该三角形是直角三角形。
解决方案：
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)
问题7：ABC是一个三角形，其中∠A= 72°。角∠B和∠C的内部等分线在O中相交。找到∠BOC的大小。
解决方案：
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°.
问题8：在任何情况下，三角形底角的等分线都不能围成直角。
解决方案：
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)
问题9：如果三角形底角的等分线围成135°角，则证明该三角形是直角三角形。
解决方案：
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)
问题10：在三角形△ABC中，, ABC =∠ACB，andABC和∠ACB的等分线在O处相交，使得∠BOC= 120°。证明∠A=∠B=∠C= 60°。
解决方案：
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)
问题11：三角形可以有吗?
(i)两个直角。
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)两个钝角
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)两个锐角
Having two acute angles does not violates any law as there sum is less than 180°
Answer: YES
(iv)所有大于60°的角度
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)所有小于60°的角度
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)所有角度等于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
问题12：如果一个三角形的每个角度都小于其他两个角度的和，请证明该三角形的所有角度均为锐角。
解决方案：
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.