Chapter 3: Problem 4
In a triangle \(\mathrm{PQR}, \angle R=\frac{\pi}{2}\). If \(\tan \left(\frac{p}{2}\right)\) and \(\tan \left(\frac{Q}{2}\right)\) are the roots of \(a x^{2}+b x+c=0, a\) \(\neq 0\), then: [AIEEE - 2005] (a) \(b=a+c\) (b) \(b=c\) (c) \(c=a+b\) (d) \(a=b+c\)
Short Answer
Expert verified
The answer is (d) \( a = b + c \).
Step by step solution
01
Identify Key Information Given in the Problem
We have a right triangle \( \triangle PQR \) where \( \angle R = \frac{\pi}{2} \). We need to determine a relationship between the coefficients \(a, b,\) and \(c\) of the quadratic equation whose roots are \(\tan\left(\frac{P}{2}\right)\) and \(\tan\left(\frac{Q}{2}\right)\).
02
Use the Half-Angle Identity for Tangent
For a right triangle, if \( \angle R = \frac{\pi}{2} \), \( P + Q = \pi/2 \). The half-angle tangent identities are: \[ \tan\left(\frac{P}{2}\right) = \sqrt{\frac{1 - \cos P}{1 + \cos P}} \] and \[ \tan\left(\frac{Q}{2}\right) = \sqrt{\frac{1 - \cos Q}{1 + \cos Q}} \].
03
Relate Angles P and Q in Terms of Triangle Sides
Using the triangle sides, \( \tan P = \frac{QR}{PR} \) and \( \tan Q = \frac{PR}{QR} \). Also, using \( P + Q = \frac{\pi}{2} \), then \( \tan Q = \cot P \). Hence, we have \( \tan Q \cdot \tan P = 1 \).
04
Quadratic Equation Using Product of Roots Identity
For roots \( \tan\left(\frac{P}{2}\right) \) and \( \tan\left(\frac{Q}{2}\right) \), the product is given by the quadratic identity: \[ \tan\left(\frac{P}{2}\right) \times \tan\left(\frac{Q}{2}\right) = \frac{1 - \frac{\cos P}{\cos Q}}{1 + \frac{\cos P}{\cos Q}} = \frac{PQ}{PR} \times \frac{PR}{PQ} = 0 \].
05
Apply Viète's Formulas
According to Viète's formulas for the quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots \(s = \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = -\frac{b}{a}\) and the product of the roots \(p = \tan\left(\frac{P}{2}\right) \cdot \tan\left(\frac{Q}{2}\right) = \frac{c}{a}\). Given the derived product in Step 4, \( p = 0 \), so \( c = 0 \). Unfortunately, since both roots are zero, this term is eliminated.
06
Interpret and Choose the Correct Option
From the product \(\tan\left(\frac{P}{2}\right) \cdot \tan\left(\frac{Q}{2}\right) = 0\), which implies \( c = 0 \), the potential relation is \(b = a\), leaving \(b = a + c\) where \(c = 0\), hence \(b = a + 0 = a\). Thus the correct choice is \( b = a + c \). Hence, the selected option is \(d.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Right Triangle Properties
A right triangle is a special kind of triangle where one angle is exactly 90 degrees or \( \frac{\pi}{2} \) radians. In such triangles, a few properties help us understand their geometry better:
- **Pythagorean Theorem**: This states that for a right triangle with sides \( a \), \( b \), and hypotenuse \( c \), the relationship is \( a^2 + b^2 = c^2 \).
- **Complementary Angles**: The two non-right angles, let's call them \( P \) and \( Q \), are complementary, meaning \( P + Q = \frac{\pi}{2} \) or \( 90^\circ \).
- **Trigonometric Ratios**: These triangles are well-known for their use in defining sine, cosine, and tangent. The tangent of an angle, like \( P \), can be defined as \( \tan P = \frac{\text{opposite}}{\text{adjacent}} \). Therefore, \( \tan P \) and \( \tan Q \) are inverses since \( \tan Q = \cot P \).
Quadratic Equations
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a eq 0 \). These are solved to find the values of \( x \) that satisfy the equation. Here are a few key points about quadratic equations:
- **Standard Form**: The typical form is \( ax^2 + bx + c = 0 \), and it's polynomial with a degree of 2.
- **Roots**: The values of \( x \) which satisfy the equation are called the roots, which might be real or complex.
- **Factoring Method**: Sometimes, quadratics can be factorized into two binomials which, when set to zero, can yield the roots.
- **Quadratic Formula**: For quadratics that are not easily factored, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used to find the roots.
Viète's Formulas
Viète's formulas provide a deep insight into the relationship between the coefficients and the roots of a polynomial equation. For a quadratic equation \( ax^2 + bx + c = 0 \):
- **Sum of Roots**: If \( \alpha \) and \( \beta \) are the roots, then their sum is given by \( \alpha + \beta = -\frac{b}{a} \).
- **Product of Roots**: The product of the roots is \( \alpha \beta = \frac{c}{a} \).