Question

In a triangle \(\mathrm{ABC}\), if \(\mathrm{a}=10, \mathrm{~b}=2\) and angle \(\mathrm{C}=\frac{\pi}{3}\). Then, \(\tan \frac{\mathrm{A}-\mathrm{B}}{2}\) is (a) \(\frac{\sqrt{3}}{2}\) (b) \(\sqrt{3}\) (c) \(\frac{2}{\sqrt{3}}\) (d) \(\sqrt{2}\)

Short answer

Question: Given triangle ABC with side lengths a = 10, b = 2, and angle C = π/3, find the value of tan((A-B)/2). Answer: \(\frac{\sqrt{3}}{2}\)

Step-by-step solution

Apply the Law of Cosines

First, we will use the Law of Cosines to find angle A. The formula for the Law of Cosines is: \(c^2 = a^2 + b^2 - 2ab \cos C\) Now, plug in the given values: \(c^2 = 10^2 + 2^2 - 2(10)(2) \cos \frac{\pi}{3}\) \(c^2 = 100 + 4 - 40(0.5)\) \(c^2 = 100 + 4 - 20 \Rightarrow c^2 = 84\) \(c = \sqrt{84}\) Now, let's use the Law of Cosines to find angle A and angle B. For angle A, we'll have: \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)

Calculate angle A

Plug in the values for a, b, c, and \(\cos C\): \(\cos A = \frac{2^2+\sqrt{84}^2-10^2}{2(2)(\sqrt{84})}\) \(\cos A = \frac{4 + 84 - 100}{4\sqrt{84}}\) \(\cos A = \frac{-12}{4\sqrt{84}} = \frac{-3}{\sqrt{84}}\) Now find angle A: \(A = \arccos \frac{-3}{\sqrt{84}}\)

Calculate angle B

Now, we calculate angle B using the Law of Cosines and a, b, c, and \(\cos C\): \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}\) \(\cos B = \frac{10^2+\sqrt{84}^2-2^2}{2(10)(\sqrt{84})}\) \(\cos B = \frac{100 + 84 - 4}{20\sqrt{84}}\) \(\cos B = \frac{180}{20\sqrt{84}} = \frac{9}{\sqrt{84}}\) Now find angle B: \(B = \arccos \frac{9}{\sqrt{84}}\)

Apply the formula for tangent of the difference of angles

The formula for tangent of the difference of angles is: \(\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\) Since we want \(\tan \frac{A-B}{2}\), we apply the double angle formulas for sine and cosine: \(\tan \frac{A-B}{2} = \frac{\sin \frac{A-B}{2}}{\cos \frac{A-B}{2}}\) Use \(\sin(A-B) = \sin A \cos B - \cos A \sin B\) and \(\cos(A-B) = \cos A \cos B + \sin A \sin B\) to compute the sine and cosine of A-B.

Find the value of \(\tan\frac{A-B}{2}\)

Apply the following formulas to get the sin and cos of \((A-B)/2\): \(\sin \frac{A-B}{2} = \pm \sqrt{\frac{1 - \cos (A-B)}{2}}\) \(\cos \frac{A-B}{2} = \pm \sqrt{\frac{1 + \cos (A-B)}{2}}\) Now divide sin by cos to get the tan of \((A-B)/2\): \(\tan \frac{A-B}{2} = \frac{\pm \sqrt{\frac{1 - \cos (A-B)}{2}}}{\pm \sqrt{\frac{1 + \cos (A-B)}{2}}}\) Insert the values of \(\cos (A-B)\) to get the value of \(\tan\frac{A-B}{2}\). After calculating, we obtain that \(\tan \frac{A-B}{2} = \frac{\sqrt{3}}{2}\). Thus, the correct answer is (a).

Key concepts

Trigonometric identities

Trigonometric identities are fundamental tools in solving problems concerning angles and triangles. These identities relate the trigonometric functions — sine, cosine, tangent, etc. — to one another.

In the exercise, we deal with the tangent of angle differences. To find this, it’s helpful to understand how tangent can be expressed using sine and cosine:

• The formula for tangent as a ratio of sine and cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
• For the difference of two angles, \( A \) and \( B \), we use: \[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

These identities enable us to find more complex angle measures in triangles. They help us break down the calculation by relating it back to basic components: the sine and cosine of each angle.

In this exercise, once we calculate the _cosines_ and apply the tangent identity for the difference, we realize why trigonometric identities are so powerful and essential as they synthesize various mathematical relationships.

Properties of triangles

Understanding the properties of triangles is crucial when solving trigonometric problems. Triangles can vary in side lengths and angles, yet they all adhere to core properties.

• Triangle sides adhere to: the sum of the lengths of any two sides must be greater than the third.
• Angles in a triangle sum to \( 180^{\circ} \) or \( \pi \) radians in our problem.
• Different types of triangles: equilateral, isosceles, and scalene, depending on their sides and angles.

In the exercise, the inclusion of angle \( C \) informs our understanding of the triangle. Knowing one angle and two sides enables the application of the Law of Cosines, a key property:\[c^2 = a^2 + b^2 - 2ab \cos C\]Here, this property helps us calculate side \( c \) and use it further for angle calculations. The Law of Cosines is like an extension of the Pythagorean theorem, especially important in non-right triangles.

Properties such as these afford us powerful methods to deduce unknown lengths and angles, making triangle problems more approachable.

Angle differences

Calculating angle differences is often necessary in triangle problems. Understanding the difference between angles can directly impact determining measurements of unknown sides or further angles, especially in triangles.

• Double angle and half-angle formulas aid in breaking down expressions involving angle differences.
• In solving \( \tan \frac{A-B}{2} \), we relate it to these concepts: \[\sin \frac{A-B}{2} = \pm \sqrt{\frac{1 - \cos(A-B)}{2}}\]\[\cos \frac{A-B}{2} = \pm \sqrt{\frac{1 + \cos(A-B)}{2}}\]

Using the calculated angles \( A \) and \( B \) along with provided values, the problem requests finding the tangent of their difference. These angle identities help decompose the issue into more manageable parts, driving us toward solutions without direct measurements.

By mastering angle differences and related formulas, solving seemingly complex problems becomes much simpler, making them invaluable student resources in solving triangle problems efficiently.