Finding Triangle Angles Given Sides and Included Angle

In this problem, we are given two sides of a triangle and the angle between them (the included angle). We need to find the measures of the other two angles. This type of problem can be solved using the fundamental laws of trigonometry applied to triangles, specifically the Law of Cosines and the Law of Sines, along with the property that the sum of angles in a triangle is 180°.

Understanding the Given Information

We are given triangle ABC with:

• Side a = \((1 + \sqrt{3})\) cm (side opposite to angle A)
• Side b = \(2\) cm (side opposite to angle B)
• Angle C = \(60\degree\) (angle opposite to side c)

Step-by-Step Solution to Find Other Angles

Step 1: Calculate Side c using the Law of Cosines

The Law of Cosines relates the sides of a triangle to one of its angles. The formula is: \(c^2 = a^2 + b^2 - 2ab \cos C\).

Substitute the given values into the formula:

\(c^2 = (1 + \sqrt{3})^2 + 2^2 - 2(1 + \sqrt{3})(2) \cos 60\degree\)

Calculate the terms:

• \((1 + \sqrt{3})^2 = 1^2 + 2(1)(\sqrt{3}) + (\sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3}\)
• \(2^2 = 4\)
• \(2(1 + \sqrt{3})(2) = 4(1 + \sqrt{3})\)
• \(\cos 60\degree = \frac{1}{2}\)

Substitute these back into the Law of Cosines equation:

\(c^2 = (4 + 2\sqrt{3}) + 4 - 4(1 + \sqrt{3}) \left(\frac{1}{2}\right)\)

\(c^2 = 4 + 2\sqrt{3} + 4 - 2(1 + \sqrt{3})\)

\(c^2 = 8 + 2\sqrt{3} - 2 - 2\sqrt{3}\)

\(c^2 = 6\)

So, \(c = \sqrt{6}\) cm.

Step 2: Calculate Angle B using the Law of Sines

The Law of Sines states that the ratio of a side length to the sine of its opposite angle is constant for all three sides and angles in a triangle: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\).

We know b, c, and angle C. We can use the relation \(\frac{b}{\sin B} = \frac{c}{\sin C}\) to find angle B.

\(\frac{2}{\sin B} = \frac{\sqrt{6}}{\sin 60\degree}\)

Substitute \(\sin 60\degree = \frac{\sqrt{3}}{2}\):

\(\frac{2}{\sin B} = \frac{\sqrt{6}}{\frac{\sqrt{3}}{2}}\)

\(\frac{2}{\sin B} = \frac{\sqrt{6} \times 2}{\sqrt{3}}\)

\(\frac{2}{\sin B} = \frac{\sqrt{2 \times 3} \times 2}{\sqrt{3}}\)

\(\frac{2}{\sin B} = \frac{\sqrt{2} \times \sqrt{3} \times 2}{\sqrt{3}}\)

\(\frac{2}{\sin B} = 2\sqrt{2}\)

Now, solve for \(\sin B\):

\(\sin B = \frac{2}{2\sqrt{2}}\)

\(\sin B = \frac{1}{\sqrt{2}}\)

The angles whose sine is \(\frac{1}{\sqrt{2}}\) are 45° and 135°. Since C = 60°, if B were 135°, then A would be \(180\degree - 60\degree - 135\degree = -15\degree\), which is not possible in a triangle. Therefore, angle B must be the acute angle.

\(B = 45\degree\)

Step 3: Calculate Angle A using the Angle Sum Property of a Triangle

The sum of the angles in any triangle is always 180°. So, \(A + B + C = 180\degree\).

We know B = 45° and C = 60°. Substitute these values:

\(A + 45\degree + 60\degree = 180\degree\)

\(A + 105\degree = 180\degree\)

Solve for A:

\(A = 180\degree - 105\degree\)

\(A = 75\degree\)

Conclusion

The other two angles in the triangle ABC are 75° and 45°.

Revision Table: Key Formulas Used

Additional Information on Solving Triangles

Solving a triangle means finding the lengths of its sides and the measures of its angles. The method used depends on the information given:

• SSS (Side-Side-Side): If all three sides are known, use the Law of Cosines to find one angle, then the Law of Sines for a second angle, and the angle sum property for the third.
• SAS (Side-Angle-Side): If two sides and the included angle are known (as in this problem), use the Law of Cosines to find the third side, then the Law of Sines for one of the remaining angles, and the angle sum property for the last angle.
• ASA (Angle-Side-Angle) or AAS (Angle-Angle-Side): If two angles and one side are known, use the angle sum property to find the third angle, then the Law of Sines to find the other two sides.
• SSA (Side-Side-Angle): If two sides and a non-included angle are known, this is the ambiguous case. There might be two possible triangles, one triangle, or no triangle. Use the Law of Sines to find the unknown angle opposite one of the given sides.