Problem 9
Question
If in a \(\Delta A B C, \cos A+2 \cos B+\cos C=2\), then \(a, b\) \(c\) are in (a) A.P. (b) G.P. (c) H.P. (d) None
Step-by-Step Solution
Verified Answer
The sides \(a\), \(b\) and \(c\) do not hold any relation. Thus, the correct answer is (d) None.
1Step 1: Recall the Properties of Triangles and Cosine
The first thing to consider is that in any triangle \(\Delta ABC\), \(a\), \(b\) and \(c\) are sides opposite to angles \(A\), \(B\), and \(C\) respectively. Moreover, the sum of the angles in a triangle is always \(\pi\) radians, which translates to \(A + B + C = \pi\).
2Step 2: Manipulate the Given Equation
We have the equation \(\cos A + 2\cos B + \cos C = 2\). Using the cosine rule and replacing \(B\) and \(C\) in terms of \(A\), we get \(\cos A + 2\cos(\pi - C - A) + \cos(\pi - A - B) = 2\). The cosine rule states that \(\cos(x - y) = \cos x \cos y + \sin x \sin y\), and \(\cos(\pi - x) = -\cos x\). So the equation becomes \(\cos A - 2(\cos A \cos C + \sin A \sin C) - (\cos A \cos B + \sin A \sin B) = 2\).
3Step 3: Simplify the Equation
Simplifying the equation further and assuming that all terms involving \(A\), \(B\), and \(C\) are zero since they are constants, the left hand side then simplifies to -2\(\cos A (\cos B + \cos C) + 2\sin A (\sin B + \sin C) = -2\cos A\). Equating it to 2 (the right-hand side), we get \(-2\cos A = 2\), consequently, \(\cos A = -1\).
4Step 4: Find the Relationship Between the Sides
Given \(\cos A = -1\), it means angle \(A = \pi\) radians or \(180^\circ\). In a triangle, if one angle is \(180^\circ\), the other two angles must be \(0^\circ\), hence \(B = C = 0^\circ\). Now, we are required to find the relationship between the sides, which can be done using the sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\). Since \(\sin 0 = 0\) and \(\sin 180 = 0\), substitute \(A = 180^\circ\), \(B = C = 0^\circ\) into this equation. We get \(\frac{a}{0} = \frac{b}{0} =\frac{c}{0}\), which is undefined, meaning the three sides do not hold any common ratio or common difference. Hence, \(a\), \(b\) and \(c\) are not in AP, GP or HP.
Key Concepts
Cosine RuleTriangle Angle SumArithmetic ProgressionGeometric ProgressionHarmonic Progression
Cosine Rule
The Cosine Rule is an essential trigonometric tool for solving triangles, especially when dealing with non-right triangles. It connects the lengths of a triangle's sides with the cosine of one of its angles. The formula is given by \(c^2 = a^2 + b^2 - 2ab \cdot \cos C\), where:
- \(a, b, \) and \(c\) are the lengths of the sides of the triangle.
- \(C\) is the angle opposite to side \(c\).
Triangle Angle Sum
In any triangle, the sum of all interior angles is always \(180^\circ\) or \(\pi\) radians. This property is crucial in solving many geometric problems. Let's break it down:
- Angle \(A + B + C = 180^\circ\).
- Knowing two angles allows you to find the third.
- This rule applies uniformly to all types of triangles – whether scalene, isosceles, or equilateral.
Arithmetic Progression
An Arithmetic Progression (AP) is a sequence where each term is obtained by adding a constant difference to its preceding term. If the sides of a triangle were to be in AP, this would mean:
- \(b - a = c - b\)
- The common difference is equal for all pairs of consecutive terms.
Geometric Progression
A Geometric Progression (GP) involves a sequence where each term is found by multiplying the previous one by a fixed, non-zero number called the common ratio. If triangle sides were considered to form a GP:
- \(\frac{b}{a} = \frac{c}{b}\)
- Every subsequent side length is a product of the previous side and the common ratio.
Harmonic Progression
In a Harmonic Progression (HP), the reciprocals of the terms form an arithmetic progression. This means that if sides \(a, b,\) and \(c\) were in HP:
- The reciprocal sequence \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) forms an AP.
- There’s a common difference between the reciprocals.
Other exercises in this chapter
Problem 8
In triangle \(A B C\), prove that, \(\frac{\text { area of the in-circle }}{\text { area of triangle } A B C}\) \(=\frac{\pi}{\cot \left(\frac{A}{2}\right) \cdo
View solution Problem 8
In a tringle \(\Delta A B C\), prove that \((a-b)^{2} \cos ^{2}\left(\frac{C}{2}\right)+(a+b)^{2} \sin ^{2}\left(\frac{C}{2}\right)=c^{2}\)
View solution Problem 9
With usual notation, if in a triangle \(A B C\), \(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}\), then prove that, \(\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\
View solution Problem 10
If the circumference of the \(\Delta A B C\) lies on its incircle, then prove that, \(\cos A+\cos B+\cos C=\sqrt{2}\)
View solution