Problem 1 - A Complete Resource Book in Mathematics for JEE Main

Question

An isosceles \(\triangle A B C\) is inscribed in a circle \(x^{2}+y^{2}=a^{2}\) with the vertex \(A\) at \((a, 0)\) and the base angles \(B\) and \(C\) each equal to \(75^{\circ}\) then length of the base \(B C\) is (A) \(\frac{a}{2}\) (B) \(a\) (C) \(\frac{2 a}{\sqrt{3}}\) (D) \(\frac{\sqrt{3} a}{2}\)

Step-by-Step Solution

Answer

The length of base BC is \( a \).

Step 1: Understanding the Problem

We are given an isosceles triangle \( \triangle ABC \) inscribed in a circle with equation \( x^2 + y^2 = a^2 \). This means the circle has center \( (0, 0) \) and radius \( a \). The vertex \( A \) of the triangle is at \((a, 0)\). The base angles \( \angle ABC \) and \( \angle ACB \) both measure \( 75^\circ \). We need to find the length of the base \( BC \).

Step 2: Use Circle Properties

Because the triangle is inscribed, the vertex \( A \) on the circle will be one point and the other two, \( B \) and \( C \), must also lie on the circle. The angle subtended by an arc at the circle's center is twice the angle subtended at the circumference. Since \( \angle BAC = 30^\circ \), \( \angle BOC = 2 \times \angle BAC = 60^\circ \).

Step 3: Determine Triangle \( OBC \)

The angle \( \angle BOC = 60^\circ \) suggests \( \triangle BOC \) is an equilateral triangle because \( OB \) and \( OC \) are both radii of the circle, each of length \( a \). Thus, \( BC = OB = OC = a \).

Step 4: Conclude with Calculation

Given that \( \triangle BOC \) is equilateral with each side equal to \( a \), the length of \( BC \) is determined directly from this. Therefore, \( BC = a \).

Key Concepts

circle properties

Understanding circle properties is essential in solving geometry problems involving circles. A circle in geometry is typically defined by its center and radius. The equation for a circle centered at the origin \(0,0\) with radius \a\ is given by \(x^2 + y^2 = a^2\). This equation describes all the points equidistant from the center with distance equal to the radius.
In our problem, the circle's center is located at the origin \(0, 0\), and its radius is \a\. Key circle properties are:

• The distance from the center to any point on the circle is constant, namely \a\.
• Every triangle inscribed in a circle with one side as the diameter is a right triangle. However, our triangle is not inscribed on the diameter.

These properties are useful when a triangle is inscribed, as each vertex must lie on the circle's circumference.

triangle geometry

Triangles are fundamental geometric shapes, and understanding their properties is vital in problem-solving. An isosceles triangle, like \(\triangle ABC\) in our problem, has two equal sides and two equal angles. Here, the equal angles are at the base, each measuring \75^\circ\.
Key facts about isosceles triangles include:

• The angles opposite the equal sides are equal.
• The line drawn from the vertex angle to the base (the altitude) not only bisects the base but also bisects the vertex angle.

In many circle-related problems, we need to consider the specific position of the triangle's vertices to use the properties effectively.

geometry problem-solving

Solving geometry problems often involves synthesizing various geometric principles. First, one must clearly understand the problem constraints, such as inscribed angles and circle properties. In our exercise, identifying the inscribed angles and recognizing that \angle BOC = 60^\circ\, a known property where the angle at the center is twice the angle at the circle's circumference, helped us find the triangle \(\triangle BOC\) as equilateral.
Effective problem-solving strategies include:

• Clearly define known quantities and relationships.
• Work through geometric properties step by step.
• Visualize with diagrams to see relationships and angles.

By using these methods, we determined that \(BC = a\) without complex calculations, understanding the mechanics behind \triangle BOC\ being equilateral.