The sides of a triangle are m, n and \(\rm \sqrt{m^2+n^2+mn}\) . What is the sum of the acute angles of the triangle?
60°
Finding the Sum of Acute Angles in the Triangle
Let the three sides of the triangle be \(a=m\), \(b=n\), and \(c=\rm \sqrt{m^2+n^2+mn}\). Let the angles opposite to these sides be A, B, and C respectively.
To find the angles of the triangle, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, c, and angles A, B, C (opposite to sides a, b, c respectively):
- \(a^2 = b^2 + c^2 - 2bc \cos A\)
- \(b^2 = a^2 + c^2 - 2ac \cos B\)
- \(c^2 = a^2 + b^2 - 2ab \cos C\)
Let's use the third formula to find angle C, which is opposite the side \(c=\rm \sqrt{m^2+n^2+mn}\).
Substituting the given side lengths into the formula:
\(c^2 = a^2 + b^2 - 2ab \cos C\)
\((\rm \sqrt{m^2+n^2+mn})^2 = m^2 + n^2 - 2mn \cos C\)
\(m^2+n^2+mn = m^2 + n^2 - 2mn \cos C\)
Now, we can simplify the equation by subtracting \(m^2 + n^2\) from both sides:
\(mn = -2mn \cos C\)
Assuming m and n are positive lengths (as they are sides of a triangle), we can divide both sides by \(mn\):
\(1 = -2 \cos C\)
\(\cos C = -\frac{1}{2}\)
The angle C whose cosine is \(-\frac{1}{2}\) is \(120^\circ\).
So, one angle of the triangle is \(120^\circ\). This is an obtuse angle (greater than \(90^\circ\)). A triangle can have at most one obtuse angle. Therefore, the other two angles, A and B, must be acute angles (less than \(90^\circ\)).
The sum of the angles in any triangle is always \(180^\circ\).
\(A + B + C = 180^\circ\)
Substitute the value of angle C:
\(A + B + 120^\circ = 180^\circ\)
To find the sum of angles A and B, subtract \(120^\circ\) from \(180^\circ\):
\(A + B = 180^\circ - 120^\circ\)
\(A + B = 60^\circ\)
The sum of the acute angles (A and B) of the triangle is \(60^\circ\).
Let's summarise the angles:
| Angle | Value | Type |
|---|---|---|
| C | \(120^\circ\) | Obtuse |
| A | Acute | Acute |
| B | Acute | Acute |
| Sum of A and B | \(60^\circ\) | Sum of Acute Angles |
Understanding Triangle Angles and Side Lengths
This problem demonstrates how the Law of Cosines relates the lengths of the sides of a triangle to the cosines of its angles. By knowing the side lengths, we can determine the nature and values of the angles.
- If \(c^2 = a^2 + b^2\), angle C is \(90^\circ\) (Right triangle).
- If \(c^2 < a^2 + b^2\), angle C is acute (\(< 90^\circ\)).
- If \(c^2 > a^2 + b^2\), angle C is obtuse (\(> 90^\circ\)).
In our case, \(c^2 = m^2+n^2+mn\). Since \(mn\) is positive (as m and n are positive lengths), \(c^2\) is greater than \(m^2+n^2\). This confirms that angle C is obtuse, as we found (\(120^\circ\)).
Revision Table: Key Geometric Concepts
| Concept | Description | Formula/Property |
|---|---|---|
| Law of Cosines | Relates the length of a side of a triangle to the lengths of the other two sides and the cosine of the angle between them. | \(c^2 = a^2 + b^2 - 2ab \cos C\) (and cyclic permutations) |
| Sum of Angles in a Triangle | The sum of the interior angles of any triangle is always \(180^\circ\). | \(A + B + C = 180^\circ\) |
| Acute Angle | An angle measuring less than \(90^\circ\). | \(0^\circ < \theta < 90^\circ\) |
| Obtuse Angle | An angle measuring greater than \(90^\circ\) but less than \(180^\circ\). | \(90^\circ < \theta < 180^\circ\) |
Additional Information on Triangle Classification by Angles
Triangles can be classified based on their angles:
- Acute Triangle: All three angles are acute (less than \(90^\circ\)).
- Right Triangle: One angle is exactly \(90^\circ\). The other two angles are acute and their sum is \(90^\circ\).
- Obtuse Triangle: One angle is obtuse (greater than \(90^\circ\)). The other two angles are acute and their sum is less than \(90^\circ\).
In this problem, since one angle is \(120^\circ\), the triangle is an obtuse triangle. Consequently, the other two angles must be acute. The sum of these two acute angles must be \(180^\circ - 120^\circ = 60^\circ\).
Similar Questions
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Which of the above statements is/are correct?
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Consider the following statements:
1. If ABC is a right-angled triangle, right-angled at A, and if sin \(\rm B = \frac 1 3,\) then cosec C = 3.
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Which of the above statements is/are correct?
- In a triangle \(ABC\), if \(a\), \(b\) and \(c\) are the lengths of the sides opposite to the angles \(A\), \(B\) and \(C\) respectively, then what is \(\frac{\sin(A-B)}{\sin(A+B)}\) equal to ?
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