Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the sides of a right angled triangle are in G.P., then the cosine of the greater acute angle is (A) \(\frac{1}{1+\sqrt{5}}\) (B) \(\frac{1}{1-\sqrt{5}}\) (C) \(\frac{1+\sqrt{5}}{\underline{\phantom{xx}}}\) (D) None of these

Step-by-Step Solution

Answer

The cosine of the greater acute angle is \(\frac{1}{1+\sqrt{5}}\) (Option A).

Step 1: Define a General Geometric Progression

The sides of a right-angled triangle are in geometric progression (G.P.), where if the sides are considered as \(a\), \(ar\), and \(ar^2\), then \(a\) serves as the smallest side, \(ar\) as the middle side, and \(ar^2\) as the largest side (the hypotenuse in the case of a right-angled triangle). Hence, the side of the triangle can be described as, \(a\), \(ar\), and \(ar^2\).

Step 2: Apply Pythagoras Theorem

For a right-angled triangle, the square of the hypotenuse \((ar^2)^2\) should equal the sum of the squares of the other two sides (\(a^2\) and \((ar)^2\)). Thus, we can write the equation: \(a^2 + (ar)^2 = (ar^2)^2\). Simplifying, \(a^2 + a^2 r^2 = a^2 r^4\). Divide the whole equation by \(a^2\) to get \(1 + r^2 = r^4\).

Step 3: Solve the Quadratic Equation

Re-arrange the equation \(1 + r^2 = r^4\) to \(r^4 - r^2 - 1 = 0\). Let \(x = r^2\), substituting gives \(x^2 - x - 1 = 0\). Solve this quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). With \(a = 1\), \(b = -1\), and \(c = -1\), we have \(x = \frac{1 \pm \sqrt{5}}{2}\). Since \(x = r^2\) cannot be negative, we take \(x = \frac{1 + \sqrt{5}}{2}\).

Step 4: Determine the Cosine of the Greater Acute Angle

In a right-angled triangle, the cosine of an angle is the ratio of the adjacent side to the hypotenuse. The greater acute angle corresponds to the side ratio \(a/ar^2\), which simplifies to \(1/r^2\). From Step 3, we found \(r^2 = \frac{1 + \sqrt{5}}{2}\), so the cosine of the greater acute angle is \(\frac{1}{\frac{1 + \sqrt{5}}{2}} = \frac{2}{1+\sqrt{5}}\). Multiply numerator and denominator by \(1-\sqrt{5}\) to rationalize: \[ \frac{2(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})} = \frac{2 - 2\sqrt{5}}{-4} = \frac{1}{1+\sqrt{5}}.\]

Step 5: Conclusion

The cosine of the greater acute angle is \(\frac{1}{1+\sqrt{5}}\) which matches option (A).

Key Concepts

Cosine of Angles

In the context of right-angled triangles, the cosine of an angle is fundamental to defining relationships between the triangle's sides. When we talk about the cosine of an angle, especially in a right-angled triangle, we are often referring to the ratio of the length of the adjacent side to the length of the hypotenuse.
This forms a critical trigonometric identity. To determine the cosine of the greater acute angle in the right triangle given in the problem, we first need to remember that the hypotenuse is the longest side. The cosine of the greater acute angle is thereby calculated using the formula:

• Cosine (greater acute angle) = Adjacent side / Hypotenuse

For the exercise involving a triangle where sides are in geometric progression, if the sides are denoted as \(a\), \(ar\), and \(ar^2\) (where \(ar^2\) is the hypotenuse), the side opposing the greater acute angle becomes \(a\), and the adjacent is \(ar\). Substituting values, we get \(\text{Cosine} = \frac{a}{ar^2} = \frac{1}{r^2}\).
Using the value of \(r^2 = \frac{1 + \sqrt{5}}{2}\), the final computation provides the solution \(\frac{1}{1+\sqrt{5}}\), aligning with option (A).

Pythagorean Theorem

The Pythagorean Theorem is a cornerstone of geometry regarding right-angled triangles. It provides a relationship between the three sides of the triangle:

• The square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

Mathematically, this is expressed as: \(c^2 = a^2 + b^2 \), where \(c\) is the hypotenuse.In this exercise, we apply the theorem to a triangle with sides in geometric progression: \(a, ar, ar^2\). Calculating based on the progression, you have \(a^2 + (ar)^2 = (ar^2)^2\). Simplifying this leads to: \[a^2 + a^2r^2 = a^2r^4\] Dividing through by \(a^2\), the equation becomes \(1 + r^2 = r^4\). This step is key to solving for \(r\), and later for finding the cosine of the involved angles, by translating these geometric terms into a recognisable quadratic equation.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(x\) represents an unknown value, and \(a\), \(b\), and \(c\) are constants. Quadratic equations can often be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In the triangle problem, after simplifying using the Pythagorean theorem, we turned to quadratic equations to find \(r^2\). Translating the equation \(1 + r^2 = r^4\) into a format suitable for our formula, it becomes \(r^4 - r^2 - 1 = 0\). Upon setting \(x = r^2\), the equation transforms to a more familiar form: \(x^2 - x - 1 = 0\).Using the quadratic formula where \(a = 1\), \(b = -1\), and \(c = -1\): \[ x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]From the solutions, \(x = \frac{1 + \sqrt{5}}{2}\) is selected as \(x\), since equivalents of \(r^2\) must be positive. This selection leads to the computation of cosine in the context of our triangle problem. Such quadratic equations often surface in trigonometry, influencing sides and angles in polygons just the same.