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

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) \(\frac{1+\sqrt{5}}{2}\) (D) None of these

Step-by-Step Solution

Answer

Option (A) \(\frac{1}{1+\sqrt{5}}\).

Step 1: Understand the Triangle and Given Condition

In a right-angled triangle with sides in a geometric progression (G.P.), denote the sides as \(a\), \(ar\), and \(ar^2\). The square of the largest side should equal the sum of the squares of the other two sides as per the Pythagorean theorem. Assume \(ar^2\) as the hypotenuse for clarification.

Step 2: Set up the Pythagorean Theorem

The Pythagorean theorem states: \((ar^2)^2 = a^2 + (ar)^2\). Expand and simplify this to \(a^2r^4 = a^2 + a^2r^2\). Factor out \(a^2\) to get \(a^2(r^4 - r^2 - 1) = 0\). Since \(a eq 0\), divide both sides by \(a^2\) to obtain \(r^4 - r^2 - 1 = 0\).

Step 3: Solve for \(r^2\)

Let \(x = r^2\). The equation becomes \(x^2 - x - 1 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = -1\). Calculate \(x = \frac{1 \pm \sqrt{1 + 4}}{2}\), leading to \(x = \frac{1 \pm \sqrt{5}}{2}\). Discard the negative value, so \(r^2 = \frac{1 + \sqrt{5}}{2}\).

Step 4: Determine Cosine of the Greater Acute Angle

The cosine of the angle opposite the side of length \(ar\) can be found using the formula \(\cos(\theta) = \frac{adjacent}{hypotenuse} = \frac{a}{ar^2} = \frac{1}{r^2}\). Substitute \(r^2\) from the previous calculation to get \(\cos(\theta) = \frac{1}{\frac{1+\sqrt{5}}{2}} = \frac{2}{1 + \sqrt{5}}\), which simplifies to \(\frac{1}{1 + \sqrt{5}}\) after rationalizing the denominator.

Key Concepts

Geometric Progression in Triangles

A geometric progression (G.P.) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the case of a right-angled triangle, if the sides are in G.P., this means:

• The first side can be represented as \( a \).
• The second side, using the common ratio \( r \), becomes \( ar \).
• The third side is \( ar^2 \).

We assume \( ar^2 \) to be the hypotenuse because it is the largest side, as is typical in a right-angled triangle.
This arrangement helps in easily applying the Pythagorean theorem, laying a foundation for further calculations in resolving angles and side lengths.

Pythagorean Theorem and its Role

The Pythagorean theorem is a fundamental principle used to solve problems related to right-angled triangles. It is expressed mathematically as:
\[ c^2 = a^2 + b^2 \]
where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides of the triangle.
When the sides are in a geometric progression, let the sides of the triangle be \( a \), \( ar \), and \( ar^2 \) with \( ar^2 \) as the hypotenuse. Plug these into the Pythagorean theorem:

• \( (ar^2)^2 = a^2 + (ar)^2 \)
• Leads to \( a^2r^4 = a^2 + a^2r^2 \)
• From this, we can derive the equation \( r^4 - r^2 - 1 = 0 \) after simplification.

This quadratic equation can be solved for \( r^2 \) to find the common ratio, which is critical for further computations in such problems.

Cosine of an Angle in a Right Triangle

The cosine of an angle in a right-angled triangle is a trigonometric function that represents the ratio of the length of the adjacent side to the hypotenuse. In this specific problem, we need to find the cosine of the angle opposite the side \( ar \):

The formula is:

• \( \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} \)
• Here, it translates to \( \cos(\theta) = \frac{a}{ar^2} \)
• This simplifies to \( \frac{1}{r^2} \)

Substituting \( r^2 = \frac{1 + \sqrt{5}}{2} \), we find:

• \( \cos(\theta) = \frac{1}{\frac{1 + \sqrt{5}}{2}} \)
• Simplifying further gives \( \cos(\theta) = \frac{2}{1 + \sqrt{5}} \)
• After rationalizing, the solution is \( \frac{1}{1 + \sqrt{5}} \)

The angle whose cosine is this value is the greater acute angle of the right-angled triangle.