Answer:
Step-by-step explanation:
Given SinP + SinQ = 7/5...1 and
∠P + ∠Q = 90°... 2
From compound angle; SinP +SinQ = [tex]2sin(\frac{P+Q}{2} )cos(\frac{P-Q}{2} )[/tex]... 3
Substituting equation 2 into 3 we will have;
SinP +SinQ = [tex]2sin(\frac{90}{2} )cos(\frac{P-Q}{2} )[/tex] = 7/5
[tex]2sin45^{0} cos\frac{P-Q}{2}=7/5[/tex]
since P = 90-Q from equation 1, then;
[tex]2sin45^{0} cos\frac{90-Q-Q}{2}=7/5\\2sin45^{0} cos\frac{90-2Q}{2}=7/5\\2(\frac{1}{\sqrt{2} } ) cos\frac{90-2Q}{2}=7/5\\cos\frac{90-2Q}{2} = 7/5* \frac{\sqrt{2} }{2} \\cos\frac{90-2Q}{2} = \frac{7\sqrt{2}}{10}\\\frac{90-2Q}{2} = cos^{-1} \frac{7\sqrt{2}}{10}\\\frac{90-2Q}{2} = 8.15\\90-2Q = 16.30\\2Q = 90-16.3\\2Q = 73.7\\Q = 36.85^{0} \\\\P = 90-36.85\\P = 53.15^{0}[/tex]
To get sin2P; Accoding to the trig identity;
Sin2P = 2SinPCosP
Sin2P = 2Sin53.15cos53.15
sin2P = 0.9598
sin2P ≈ 1
