Let P Q R be a triangle of area Δ with a=2, b=7 / 2, and c=5 / 2, where a, b, and c are the lengths of the sides of the triangle opposite to the angles at P, Q, and R, respectively. Then (2 sin P- sin 2 P/2 sin P+ sin 2 P) equals

Question

Let P Q R be a triangle of area Δ with a=2, b=7 / 2, and c=5 / 2, where a, b, and c are the lengths of the sides of the triangle opposite to the angles at P, Q, and R, respectively. Then (2 sin P- sin 2 P/2 sin P+ sin 2 P) equals (A) (3/4 Δ) (B) (45/4 Δ) (C) ((3/4 Δ))2 (D) ((45/4 Δ))2

Tardigrade Admin · Accepted Answer

(2 sin P-2 sin P cos P/2 sin P+2 sin P cos P) =(1- cos P/1+ cos P)=(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))2/Δ2) =((((1/2))((3/2)))2/Δ2) =((3/4 Δ))2

Q. Let $PQR$ be a triangle of area $Δ$ with $a = 2, b = 7/2$ , and $c = 5/2$ , where $a, b$ , and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ , and $R$ , respectively. Then $\frac{2 s i n P - s i n 2 P}{2 s i n P + s i n 2 P}$ equals

$\frac{3}{4Δ}$

$\frac{45}{4Δ}$

$(\frac{3}{4Δ})^{2}$

$(\frac{45}{4Δ})^{2}$

Q. Let PQR be a triangle of area Δ with a=2,b=7/2, and c=5/2, where a,b, and c are the lengths of the sides of the triangle opposite to the angles at P,Q, and R, respectively. Then 2sinP+sin2P2sinP−sin2P​ equals

4Δ3​

4Δ45​

(4Δ3​)2

(4Δ45​)2

2sinP+2sinPcosP2sinP−2sinPcosP​ =1+cosP1−cosP​=2cos22P​2sin22P​​ =tan22P​ =s(s−a)(s−b)(s−c)​ =Δ2((s−b)(s−c))2​ =Δ2((21​)(23​))2​ =(4Δ3​)2

Q. Let $PQR$ be a triangle of area $Δ$ with $a = 2, b = 7/2$ , and $c = 5/2$ , where $a, b$ , and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ , and $R$ , respectively. Then $\frac{2 s i n P - s i n 2 P}{2 s i n P + s i n 2 P}$ equals

$\frac{3}{4Δ}$

$\frac{45}{4Δ}$

$(\frac{3}{4Δ})^{2}$

$(\frac{45}{4Δ})^{2}$