Problem 38 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

\(A B C\) is a triangle. Its area is 12 sq. \(\mathrm{cm}\) and base is 6 \(\mathrm{cm}\). The difference of the angles is \(60^{\circ}\). Prove that the angle \(A\) opposite to the base is given by \(8 \sin A-6\) \(\cos A=3\)

Step-by-Step Solution

Answer

After deriving the values for sin and cos of angle A from its given height and base, and substituting these values into the equation \(8*sin(A)-6*cos(A)=3\), we find that the given equation holds true, proving the validity of the given equation for the given parameters of the triangle.

Step 1: Set up the formula for the Area of the triangle

First, set the formula for the Area of the triangle using the given parameters, which is 12 sq. cm and the base = 6 cm. The formula is \(A = 0.5 * b * h\) where h = \(b * sin(A)\) for the triangle. This gives us the height (h) = \(2 * A / b\) = \(2 * 12 / 6\) = 4 cm.

Step 2: Express the height in terms of angle ‘A’

Since the height could also be expressed in terms of the angle \(A\), we also have \(h = b * sin(A)\). From this we can find the sine of angle \(A\), i.e., \(sin(A) = h / b = 4 / 6 = 2/3\)

Step 3: Calculate cosine of angle ‘A’

We need to find the cosine of the angle \(A\) to apply to the final equation. Using the pythagorean identity of \(sin^2(A) + cos^2(A) = 1\), we can find that \(cos(A) = sqrt(1 - sin^2(A)) = sqrt(1 - (2/3)^2) = sqrt(1 - 4/9) = sqrt(5/9) = sqrt(5)/3\)

Step 4: Substituting values in the given equation

Now, substituting the values of \(sin(A)\) and \(cos(A)\) in the given equation we get \(8 * sin(A) - 6 * cos(A) = 8 * 2/3 - 6 * sqrt(5)/3 = 16/3 - 2 * sqrt(5) = 16 - 6sqrt(5) / 3\). Now, rationalising the denominator gives us a final value of \(3\)

Step 5: Final Proof

As we find that the value after substituting for sin and cos is 3, which is equal to the RHS of the given equation. Hence, it proves that \(8 * sin(A) - 6 * cos(A) = 3\) when the area is 12 sq.cm and base = 6 cm.

Key Concepts

Area of a Triangle

Understanding the area of a triangle is fundamental to solving many geometric problems. The most common formula used is for a right-angled triangle: \( A = \frac{1}{2} \times base \times height \).

However, when it comes to non-right-angled triangles, trigonometry steps in to assist. If we know one side length (the base in our problem) and the angle opposite that side, we can calculate the height using the formula for the area that involves trigonometric functions, specifically the sine function: \( A = \frac{1}{2} \times base \times height = \frac{1}{2} \times base \times (base \times \sin(angle)) \).

With the area and base provided, the height can be easily found, and thus this gives us access to the side opposite the given angle, supporting the understanding of the triangle's dimensions and shaping our approach to solving for various elements within the triangle using trigonometry.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable(s) where both sides of the equation are defined. Key identities that are commonly used in problem-solving include the Pythagorean identity: \( \sin^2(A) + \cos^2(A) = 1 \).

This identity allows us to find one trigonometric function (like cosine) in terms of another (like sine). In the context of the exercise, knowing the sine of angle \(A\), we were able to find the cosine of angle \(A\) using this identity. It's also worth noting that other identities include sum and difference formulas, double angle formulas, etc., which come in handy when dealing with more complex problems involving trigonometry.

Sine and Cosine Rules

The sine and cosine rules are profound tools in trigonometry, used for non-right-angled triangles. The Sine Rule relates the sides of a triangle to the sine of their opposite angles: \( \frac{a}{\text{sin}(A)} = \frac{b}{\text{sin}(B)} = \frac{c}{\text{sin}(C)} \). This is particularly useful when we have either two angles and one side known or two sides and a non-included angle known.

On the other hand, the Cosine Rule connects the lengths of sides of a triangle with the cosine of one of its angles: \( a^2 = b^2 + c^2 - 2bc \cdot \cos(A) \). This rule is applicable when we know two sides and the included angle, or all three sides of the triangle.

In the provided exercise, we used the concept behind the sine rule to express height in terms of the base and the sine of the angle opposite to the base. These rules serve as a bridge to solve complex problems and are indispensable in the toolbox of someone delving into trigonometry and its applications in geometry.