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Mathematics Question on Three Dimensional Geometry

If PQR is a triangle of area △\triangle with a = 2, b = 72 \frac{7}{2} and c = 52 \frac{5}{2}, where a, b and c are the lengths of the sides of ' the triangle opposite to the angles a t P ,Q and R, respectively. Then, 2 sin P−sin 2P2 sin P+sin 2P \frac{ 2 \, sin \, P - sin \, 2 P }{ 2 \, sin \, P + sin \, 2P } equals

A

34△\frac{ 3}{4 \triangle}

B

454△ \frac{ 45}{4 \triangle}

C

(34△)2\bigg( \frac{ 3}{4 \triangle} \bigg)^2

D

(454△)5\bigg(\frac{ 45}{4 \triangle}\bigg)^5

Answer

(34△)2\bigg( \frac{ 3}{4 \triangle} \bigg)^2

Explanation

Solution

PLAN If △\triangle ABC has sides a, b, c Then, tan (A/2) = (s−b) (s−a)s (s−a)\sqrt{ \frac{ (s - b) \, (s - a)}{ s \, (s - a)}} where, s = a+b+c2⇒s=2+72+522=4\frac{a + b + c}{ 2} \Rightarrow s = \frac{ 2 + \frac{ 7}{2} + \frac{5}{2} }{2} = 4 ∴2 sin P−sin 2P2 sin P+sin 2P=2sin P(1−cos P)2 sin P(1+cos P) \therefore \frac{ 2 \, sin \, P - sin \, 2 P }{ 2 \, sin \, P + sin \, 2P } = \frac{ 2 sin \, P (1 - cos \, P )}{ 2 \, sin \, P (1 + cos \, P )} \hspace26mm = 2 sin2(P/2)2 cos2(P/2)=tan2 (P/2) \frac{ 2 \, sin^2 (P/2)}{2 \, cos^2 (P/2)} = tan^2 \, (P/2) ⇒(s−b) (s−c)s (s−a)×(s−b) (s−c)(s−b) (s−c)\Rightarrow \frac{ (s - b) \, (s - c)}{ s \, (s - a)} \times \frac{ (s- b) \, (s - c)}{ (s - b) \, (s - c)} = [(s−b)2 (s−c)2]△2=(4−72)2 (4−52)2△2=(34△)2\frac{ [ (s - b)^2 \, (s - c)^2 ] }{ \triangle^2 } = \frac{\bigg( 4 - \frac{7}{2}\bigg)^2 \, \bigg( 4 - \frac{5}{2}\bigg)^2 }{ \triangle^2 } = \bigg( \frac{ 3}{ 4 \triangle } \bigg)^2