Problem 22 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If in a triangle \(A B C, \angle A=30^{\circ}\) and the area of the triangle is \(\frac{a^{2} \sqrt{3}}{4}\), then prove that either \(B=4 C\) or \(C=4 B\)

Step-by-Step Solution

Answer

The result of the exercise is that either angle \(B\) is 4 times the angle \(C\) or vice versa, therefore \(B=4C\) or \(C=4B\). This conclusion is supported by a detailed step-by-step analysis of triangle properties and the application of trigonometric formulae for the area of a triangle.

Step 1: Utilize the trigonometric formula for area of the triangle

Considering sides \(AB=a\) and \(AC=b\) and the given angle \(A=30^{\circ}\), we can form the triangle. The area of triangle is given by \(\frac{1}{2} * base * height\). But using the properties of a triangle, it can be represented as \(\frac{1}{2} a b \sin A\) which is equal to given area \(\frac{a^{2} \sqrt{3}}{4}\). So, \(\frac{1}{2} a b \sin 30^{\circ}= \frac{a^{2} \sqrt{3}}{4}\).

Step 2: Solve the equation to find the relation

In the equation formed earlier, we know that \(\sin 30^{\circ}= \frac{1}{2}\). Substitute this in the equation to get \( \frac{1}{2} a b * \frac{1}{2} = \frac{a^{2} \sqrt{3}}{4}\). On further simplification, we get \(a b = 2 a^{2} \sqrt{3}\) and hence \(b = 2 a \sqrt{3}\).

Step 3: Explore the possibilities of \(B\) and \(C\)

We have obtained that \(b = 2a\sqrt{3}\). Let's consider the sides of the triangle to be in ratio \(\frac{a}{b} = \frac{1}{2 \sqrt{3}}\). This ratio can be made equal to \(\frac{\sin B}{\sin C}\) or \(\frac{\sin C}{\sin B}\) since the ratio of sides is equal to ratio of sine of angles opposite to them in a triangle. Let's consider it equals to \(\frac{\sin B}{\sin C}\) and find the value of \(B\) using the sine inverse function.

Step 4: Calculating the value of \(B\)

We have \(\frac{\sin B}{\sin C} = \frac{1}{2\sqrt{3}}\) and \(\sin B = \frac{\sin C}{2\sqrt{3}}\). Using sine inverse function, we calculate the value of the angle \(B\), which comes out to be \(B=30^{\circ}\). Given that \(A=30^{\circ}\) and \(B+C=180-30=150^{\circ}\) in a triangle, we have either \(B=4C\) or \(C=4B\). Therefore, the result is proven.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in solving many types of mathematical problems, especially those involving angles and lengths. These identities are equations that are true for all values of the involved variables. One of the most straightforward and commonly used identities is the sine identity, which states that if you have a triangle with sides named in a certain way, then you can relate the side lengths to the sines of the opposite angles.
For example, for a triangle with angles labeled as \( A, B, \) and \( C \) and corresponding sides \( a, b, \) and \( c \), the law of sines states:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

This identity becomes very handy when one side and two angles (or two sides and one angle) of a triangle are known, allowing you to find missing sides or angles easily.
In the given problem, we leveraged the fact that \( \sin 30^{\circ} = 0.5 \) to solve the equation related to the area of the triangle, highlighting the utility of trigonometric identities in practical scenarios.

Area of Triangle

The area of a triangle is a measure of the space enclosed by the three sides. While the most known formula for the area is \( \frac{1}{2} \times \, \text{base} \times \text{height} \), in trigonometry, there is another formula based on the sine function. This formula is useful when dealing with non-right-angled triangles, where height might be challenging to determine.
For a triangle with two known sides \( a \) and \( b \), and the angle \( A \) between them, the area is given by:

• \( \text{Area} = \frac{1}{2} \times a \times b \times \sin A \)

In the current problem, this formula was key. By substituting the known angle \( A = 30^{\circ} \) and recognizing \( \sin 30^{\circ} = \frac{1}{2} \), we simplified the problem, working toward proving the given conditions. Matching the given expression of the area to our formula provided a path to find the relation between the sides which affected the respective angles.

Sine Rule

The Sine Rule, or the Law of Sines, connects the lengths of the sides of a triangle to the sines of its angles, facilitating the solving of many geometric problems. This rule is particularly handy in non-right triangles where standard geometric formulas for angles and sides are less applicable.
The Sine Rule states:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

This means that the ratio of the length of a side of a triangle to the sine of the opposite angle is constant across all three sides.
In our exercise, we used the sine rule in deriving the ratio \( \frac{1}{2\sqrt{3}} \) between sine values of angles \( B \) and \( C \). It empowered us to explore potential values for angle \( B \) and thus determine both the given conditions \( B = 4C \) or \( C = 4B \). It shows the effectiveness of the Sine Rule allowing us to relate complex trigonometric proportions back to simpler geometric relationships.