In a triangle $ABC$, $BC = 7$, $AC = 8$, $AB = \alpha \in \mathbb{N}$ and $\cos A = \frac{2}{3}$. If \[ 49 \cos(3C) + 42 = \frac{m}{n}, \] where $\gcd(m, n) = 1$, then $m + n$ is equal to ________.
Correct Answer: 39
Approach Solution - 1
The problem provides the lengths of two sides of a triangle \(ABC\) as \(BC=7\) and \(AC=8\), and the cosine of the included angle \(\cos A = \frac{2}{3}\). The third side \(AB = \alpha\) is a natural number. We are asked to find the value of \(m+n\), where \( \frac{m}{n} \) is the simplified fractional value of the expression \( 49 \cos(3C) + 42 \).
Concept Used:
The solution requires the application of fundamental trigonometric rules for a triangle and a trigonometric identity.
- Law of Cosines: In a triangle with sides \(a, b, c\) and opposite angles \(A, B, C\), the following relations hold: \[ a^2 = b^2 + c^2 - 2bc \cos A \] \[ c^2 = a^2 + b^2 - 2ab \cos C \]
- Triple Angle Identity for Cosine: The cosine of a triple angle can be expressed in terms of the cosine of the angle itself: \[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]
Step-by-Step Solution:
Step 1: Determine the length of the side \(AB = \alpha\).
We are given sides \(a = BC = 7\), \(b = AC = 8\), and \(c = AB = \alpha\). The angle \(A\) has \(\cos A = \frac{2}{3}\). Using the Law of Cosines for angle \(A\):
\[ a^2 = b^2 + c^2 - 2bc \cos A \]Substituting the given values:
\[ 7^2 = 8^2 + \alpha^2 - 2(8)(\alpha)\left(\frac{2}{3}\right) \] \[ 49 = 64 + \alpha^2 - \frac{32\alpha}{3} \]Rearranging the terms to form a quadratic equation in \(\alpha\):
\[ \alpha^2 - \frac{32\alpha}{3} + 15 = 0 \]Multiplying by 3 to clear the fraction:
\[ 3\alpha^2 - 32\alpha + 45 = 0 \]We solve this quadratic equation for \(\alpha\):
\[ (3\alpha - 5)(\alpha - 9) = 0 \]The possible values for \(\alpha\) are \(\alpha = \frac{5}{3}\) and \(\alpha = 9\). Since the problem states that \(\alpha \in \mathbb{N}\) (a natural number), we must choose \(\alpha = 9\). So, the length of side \(AB\) is 9.
Step 2: Calculate the value of \(\cos C\).
Now that we know all three sides of the triangle (\(a=7, b=8, c=9\)), we can use the Law of Cosines again to find \(\cos C\):
\[ c^2 = a^2 + b^2 - 2ab \cos C \]Substituting the side lengths:
\[ 9^2 = 7^2 + 8^2 - 2(7)(8) \cos C \] \[ 81 = 49 + 64 - 112 \cos C \] \[ 81 = 113 - 112 \cos C \] \[ 112 \cos C = 113 - 81 = 32 \] \[ \cos C = \frac{32}{112} = \frac{2 \times 16}{7 \times 16} = \frac{2}{7} \]Step 3: Compute the value of \(\cos(3C)\).
Using the triple angle identity \(\cos(3C) = 4\cos^3(C) - 3\cos(C)\) and substituting \(\cos C = \frac{2}{7}\):
\[ \cos(3C) = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right) \] \[ \cos(3C) = 4\left(\frac{8}{343}\right) - \frac{6}{7} \] \[ \cos(3C) = \frac{32}{343} - \frac{6 \times 49}{7 \times 49} = \frac{32}{343} - \frac{294}{343} \] \[ \cos(3C) = \frac{32 - 294}{343} = -\frac{262}{343} \]Step 4: Evaluate the given expression.
We need to find the value of \( 49 \cos(3C) + 42 \).
\[ 49 \left(-\frac{262}{343}\right) + 42 \]Since \(343 = 7^3 = 49 \times 7\), we can simplify the expression:
\[ \frac{49}{49 \times 7} \times (-262) + 42 = -\frac{262}{7} + 42 \]To combine these terms, we find a common denominator:
\[ -\frac{262}{7} + \frac{42 \times 7}{7} = \frac{-262 + 294}{7} = \frac{32}{7} \]Step 5: Find the value of \(m+n\).
The expression evaluates to \(\frac{32}{7}\). We are given that this is equal to \(\frac{m}{n}\), where \(\gcd(m, n) = 1\). Here, \(m=32\) and \(n=7\). The greatest common divisor of 32 and 7 is 1. Therefore, the values are correct.
The required sum is \(m+n\):
\[ m + n = 32 + 7 = 39 \]The final answer is 39.
Approach Solution -2
Using the cosine rule for \( \cos A \):
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc}\]
Substitute \( b = 8 \), \( c = 7 \), and \( \cos A = \frac{2}{3} \):
\[\frac{2}{3} = \frac{8^2 + 7^2 - a^2}{2 \times 8 \times 7}\]
\[\implies a^2 = 9\]
\[\implies a = 3\]
Now, calculate \( \cos C \) using the cosine rule:
\[\cos C = \frac{7^2 + 8^2 - 9^2}{2 \times 7 \times 8} = \frac{2}{7}\]
Then, for \( \cos(3C) \), we use the triple angle formula:
\[49 \cos(3C) + 42 = 49 \left( 4 \cos^3 C - 3 \cos C \right) + 42\]
Substituting \( \cos C = \frac{2}{7} \):
\[= 49 \left( 4 \left( \frac{2}{7} \right)^3 - 3 \cdot \frac{2}{7} \right) + 42\]
\[= 49 \left( \frac{32}{343} - \frac{6}{7} \right) + 42\]
\[= \frac{32}{7} + 42\]
Thus, \( m = 32 \) and \( n = 7 \), so:
\[m + n = 32 + 7 = 39\]
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Triangles Questions
- If the orthocentre of the triangle formed by the lines 2x + 3y – 1 = 0, x + 2y – 1 = 0 and ax + by – 1 = 0, is the centroid of another triangle, whose circumecentre and orthocentre respectively are (3, 4) and (–6, –8), then the value of |a– b| is_____.
- Consider a triangle \( \triangle ABC \) having the vertices \( A(1, 2) \), \( B(\alpha, \beta) \), and \( C(\gamma, \delta) \) and angles \( \angle ABC = \frac{\pi}{6} \) and \( \angle BAC = \frac{2\pi}{3} \). If the points \( B \) and \( C \) lie on the line \( y = x + 4 \), then \( \alpha^2 + \gamma^2 \) is equal to \( \dots \).
Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then:
- Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line \[\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{-1}\]and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$, then $K$ is equal to:
- If P(6, 1) be the orthocentre of the triangle whose vertices are A(5, –2), B(8, 3) and C(h, k), then the point C lies on the circle.
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,