In triangle ABC,with usual notations,if a cos | vedclass

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(B) Given $a \cos B = b \cos A$. Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$. Substituting these,we get $2R \sin A \cos B = 2R \sin B \cos

Vedclass · Accepted Answer

$\frac{c}{4} \sqrt{4a^2 - c^2}$

Vedclass · Answer

(B) Given $a \cos B = b \cos A$. Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$. Substituting these,we get $2R \sin A \cos B = 2R \sin B \cos A$,which implies $\sin A \cos B - \cos A \sin B = 0$,so $\sin(A - B) = 0$. Since $A$ and $B$ are angles of a triangle,$A = B$,meaning the triangle is isosceles with $a = b$. However,the condition $a \cos C 
eq c \cos A$ implies $\sin A \cos C 
eq \sin C \cos A$,so $\sin(A - C) 
eq 0$,meaning $A 
eq C$. Thus,the triangle is isosceles with $a = b$ and $a 
eq c$. The area of an isosceles triangle with sides $a, a, c$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. The base is $c$. The height $h$ is $\sqrt{a^2 - (c/2)^2} = \frac{1}{2} \sqrt{4a^2 - c^2}$. Area $= \frac{1}{2} 	imes c 	imes \frac{1}{2} \sqrt{4a^2 - c^2} = \frac{c}{4} \sqrt{4a^2 - c^2}$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

In $\triangle ABC$,with usual notations,if $a \cos B = b \cos A$ and $a \cos C \neq c \cos A$,then the area of $\triangle ABC$ is . . . . . . sq. units.

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