Triangle of given area

Geometry Level 3

Let $PQR$ be a triangle of area $\Delta$ with $a = 2, b = \frac{7}{2}$ and $c = \frac{5}{2}$ , 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 $\frac{ 2 sin P - sin 2 P}{ 2 sin P + sin 2 P}$ equals

\frac{45}{4\Delta}

\frac{3}{4\Delta}

\left( \frac{45}{4\Delta} \right)^2

\left( \frac{3}{4\Delta} \right)^2

1 solution

Ahmad Saad
Jan 8, 2017

s = (a+b+c)/2 = 4 ---> (s-a)=2 , (s-b)=0.5 , (s-c)=1.5

Δ = sqrt(s(s-a)(s-b)(s-c)) = sqrt6

cosP = (b^2 + c^2 - a^2)/(2bc) = 29/35

[2sinP - sin2P]/[2sinP + 2sin2P] = [1 - cosP]/[1 + cosP] = 3/32

= 9/96 = (9/{16*6}) = (3/4sqrt6)^2 = (3/4Δ)^2

Triangle of given area

1 solution

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