Problem 1 - A Complete Resource Book in Mathematics for JEE Main 2017

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

\(BC = \frac{2a}{\sqrt{3}}\), so the answer is (C).

Step 1: Identify the Points of the Triangle

The triangle is isosceles with vertex \(A\) at \((a, 0)\). It's inscribed in a circle with the equation \(x^2 + y^2 = a^2\), having the center at the origin \((0, 0)\) and radius \(a\). Thus, points \(B\) and \(C\) must be on this circle.

Step 2: Calculate the Angles

The triangle \(\triangle ABC\) is isosceles with the base angles \(\angle ABC = \angle ACB = 75^\circ\). The vertex angle \(\angle BAC\) is calculated as follows: \(180^\circ - 2 \times 75^\circ = 30^\circ\).

Step 3: Apply Inscribed Angle Theorem

The arc \(BC\) subtends the central angle \(2\theta\) where \(\theta = 75^\circ\). Therefore, the central angle subtended by the arc \(BC\) is \(2 \times 75^\circ = 150^\circ\).

Step 4: Calculate Arc Length Using Trigonometry

The arc length between points \(B\) and \(C\) corresponds to \(150^\circ\) of the circle. It's given through the equation \(\ell = \theta \times r\) where \(\ell\) is the arc length, \(\theta\) is in radians, and \(r\) is the radius. As \(150^\circ\) converts to \(\frac{5\pi}{6}\) radians, \(BC = a \cdot \angle BAC\) implies \(BC = 2a \sin 75^\circ\).

Step 5: Calculate Base Length \(BC\)

To find \(BC\), use the angle \(\angle BAC = 30^\circ\) and apply the formula using the sine function for an arc: \[ BC = 2a \sin\frac{30^\circ}{2} = 2a \sin 15^\circ = 2a \cdot \frac{(\sqrt{3} - 1)}{2\sqrt{2}}. \] Simplifying gives \(BC = \frac{2a}{\sqrt{3}}\).

Step 6: Choose the Correct Answer

The calculated length of base \(BC\) is \(\frac{2a}{\sqrt{3}}\). This corresponds to option (C).

Key Concepts

Inscribed Circle

An inscribed circle, or incircle, of a triangle is the largest circle that fits inside the triangle and is tangent to all three sides. When we talk about an isosceles triangle being inscribed in a circle, we refer to the circumcircle. This circle passes through all three vertices of the triangle.For our exercise, the isosceles triangle \(\triangle ABC\) is inscribed in the circle with the equation \(x^2 + y^2 = a^2\). This means the vertices \(A\), \(B\), and \(C\) lie on this circle. The center of this circle is at the origin \((0, 0)\), and the radius is \(a\). Inscription ensures that all properties of the circle can be used to solve problems about the triangle, such as calculating sides and angles.

Base Angles

In an isosceles triangle, two sides are equal, and the angles opposite these sides are called the base angles. These angles have equal measures in an isosceles triangle. In our particular case, the triangle \(\triangle ABC\) has two base angles, \(\angle ABC\) and \(\angle ACB\), each measuring \(75^\circ\).These angles play a crucial role in calculating other properties of the triangle, such as the length of the base \(BC\). By knowing the measure of these angles, we can determine the vertex angle, \(\angle BAC\), using the angle sum property of triangles:

• Sum of all angles in a triangle is always \(180^\circ\).
• Therefore, \(\angle BAC = 180^\circ - 2 \times 75^\circ = 30^\circ\).

With this information, further computations about the structure of the triangle become possible.

Arc Length

Arc length refers to the distance along the curved path of the circle between two points. In a circle, the arc length can be calculated using the formula: \(\ell = \theta \times r\), where \(\theta\) is the angle in radians, and \(r\) is the radius of the circle.For \(\triangle ABC\) inscribed in the circle, the base \(BC\) subtends an arc. Since the triangle has base angles of \(75^\circ\), the arc \(BC\) subtends a central angle of \(150^\circ\) (calculated as \(2 \times 75^\circ\)). Converting to radians gives us \(\frac{5\pi}{6}\). Therefore, knowing the radius \(a\), the arc length can help in determining the actual straight-line distance of \(BC\) within the circle.

Trigonometry

Trigonometry is a branch of mathematics that studies the relationships between sides and angles in triangles. In solving problems involving isosceles triangles inscribed in circles, trigonometry is an essential tool.In this case, we specifically used the sine function:

• Sine of an angle in a right-angled triangle gives the ratio of the opposite side to the hypotenuse.

Given \(\angle BAC = 30^\circ\), we can use trigonometry to express \( BC \) in terms of \( \sin 15^\circ \). The sine rule and properties of triangle angles let us find the expression \(BC = 2a \cdot \sin 15^\circ\), simplifying to \(\frac{2a}{\sqrt{3}}\).This process highlights how trigonometry connects angles with side lengths in triangle problems, especially within circles.