Question:
In a triangle \(ABC\), if
\[
a\cos A=b\cos B
\]
where \(a\neq b\), then \(\triangle ABC\) is
In a triangle \(ABC\), if
\[
a\cos A=b\cos B
\]
where \(a\neq b\), then \(\triangle ABC\) is
Show Hint
Whenever expressions involve
\[
a\cos A,\quad b\cos B,
\]
use cosine rule identities to convert them into algebraic equations in the sides.
Updated On: Jun 26, 2026
- Obtuse-angled triangle
- Equilateral triangle
- Acute-angled triangle
- Right angled triangle
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The Correct Option is D
Solution and Explanation
Step 1: Use the cosine rule for \(\cos A\).
In a triangle, \[ \cos A=\frac{b^2+c^2-a^2}{2bc} \] Therefore, \[ a\cos A = a\left(\frac{b^2+c^2-a^2}{2bc}\right) \] \[ = \frac{a(b^2+c^2-a^2)}{2bc} \]
Step 2: Use the cosine rule for \(\cos B\).
Similarly, \[ \cos B=\frac{a^2+c^2-b^2}{2ac} \] Thus, \[ b\cos B = b\left(\frac{a^2+c^2-b^2}{2ac}\right) \] \[ = \frac{b(a^2+c^2-b^2)}{2ac} \]
Step 3: Use the given condition.
Given, \[ a\cos A=b\cos B \] Substituting the obtained values, \[ \frac{a(b^2+c^2-a^2)}{2bc} = \frac{b(a^2+c^2-b^2)}{2ac} \]
Step 4: Cross multiply.
Multiplying both sides by \[ 2abc, \] we get \[ a^2(b^2+c^2-a^2) = b^2(a^2+c^2-b^2) \]
Step 5: Expand and simplify.
Expanding, \[ a^2b^2+a^2c^2-a^4 = a^2b^2+b^2c^2-b^4 \] Cancelling \[ a^2b^2, \] we get \[ a^2c^2-a^4 = b^2c^2-b^4 \] Rearranging, \[ a^2(c^2-a^2)=b^2(c^2-b^2) \] \[ (a^2-b^2)c^2=(a^2-b^2)(a^2+b^2) \]
Step 6: Use \(a\neq b\).
Since \[ a\neq b, \] we have \[ a^2-b^2\neq 0 \] Therefore, \[ c^2=a^2+b^2 \]
Step 7: Conclude the nature of the triangle.
Using the converse of Pythagoras theorem, \[ c^2=a^2+b^2 \] implies that the triangle is right angled.
Hence, \[ \boxed{\text{Right angled triangle}} \]
In a triangle, \[ \cos A=\frac{b^2+c^2-a^2}{2bc} \] Therefore, \[ a\cos A = a\left(\frac{b^2+c^2-a^2}{2bc}\right) \] \[ = \frac{a(b^2+c^2-a^2)}{2bc} \]
Step 2: Use the cosine rule for \(\cos B\).
Similarly, \[ \cos B=\frac{a^2+c^2-b^2}{2ac} \] Thus, \[ b\cos B = b\left(\frac{a^2+c^2-b^2}{2ac}\right) \] \[ = \frac{b(a^2+c^2-b^2)}{2ac} \]
Step 3: Use the given condition.
Given, \[ a\cos A=b\cos B \] Substituting the obtained values, \[ \frac{a(b^2+c^2-a^2)}{2bc} = \frac{b(a^2+c^2-b^2)}{2ac} \]
Step 4: Cross multiply.
Multiplying both sides by \[ 2abc, \] we get \[ a^2(b^2+c^2-a^2) = b^2(a^2+c^2-b^2) \]
Step 5: Expand and simplify.
Expanding, \[ a^2b^2+a^2c^2-a^4 = a^2b^2+b^2c^2-b^4 \] Cancelling \[ a^2b^2, \] we get \[ a^2c^2-a^4 = b^2c^2-b^4 \] Rearranging, \[ a^2(c^2-a^2)=b^2(c^2-b^2) \] \[ (a^2-b^2)c^2=(a^2-b^2)(a^2+b^2) \]
Step 6: Use \(a\neq b\).
Since \[ a\neq b, \] we have \[ a^2-b^2\neq 0 \] Therefore, \[ c^2=a^2+b^2 \]
Step 7: Conclude the nature of the triangle.
Using the converse of Pythagoras theorem, \[ c^2=a^2+b^2 \] implies that the triangle is right angled.
Hence, \[ \boxed{\text{Right angled triangle}} \]
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