If sin(A-B)=3/5, a/b=2 and A-B<90°, match the following. List I List II (A) tan((A-B)/2). (P) 0 (B) cot(C/2). (Q) 1/3 (C) Area Δ of triangle ABC. (R) 1 (D) r+c-s, where s is semiperimeter. (S) b 2

Mathematics - Properties of Triangle Question with Solution | TestHub

MathematicsProperties of TriangleGeneral (basic formula based)Medium2 minQB

MathematicsMediummatching list

If sin(A-B)=3/5, a/b=2 and A-B<90°, match the following.

List I	List II
(A) tan((A-B)/2).	(P) 0
(B) cot(C/2).	(Q) 1/3
(C) Area $Δ$ of triangle ABC.	(R) 1
(D) r+c-s, where s is semiperimeter.	(S) $b^{2}$

Options:

Answer:

Solution:

Since A-B<90° and sin(A-B)=3/5, tan((A-B)/2)=1/3. Napier’s rule gives cot(C/2)=1, hence C=90°. Since C=90° and a=2b, the area is ab/2= $b^{2}$ . Also r=s-c in a right triangle, so r+c-s=0.

Stream:JEESubject:MathematicsTopic:Properties of TriangleSubtopic:General (basic formula based)

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