Question

In triangle PQR,angle Q=90,angleP=theta,angle R=phi,PR=x+2,QR=x...find(1)underoot x+1cot phi (2)cos theta

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ 1. \: \sqrt{x + 1} .cot \: Φ \\ 2. \: \sqrt{x {}^{3} + x {}^{2} } .tan \: θ \\ 3. \: cos \: θ \\ \\ given : \\ ∠PQR = 90 \\ ∠RPQ = θ \\ ∠QRP = Φ$

$so \: in \: △PQR \\ ∠PQR = 90 \\ by \: pythagoras \: theorem \\ we \: get \\ \\ PQ {}^{2} + QR {}^{2} = PR {}^{2} \\ \\ x {}^{2} + PQ{}^{2} = (x + 2) {}^{2} \\ \\ PQ{}^{2} = (x + 2) {}^{2} - x {}^{2} \\ \\ = x {}^{2} + 4x + 4 - x {}^{2} \\ \\ = 4x + 4 \\ \\ = 4(x + 1) \\ \\ PQ= 2. \sqrt{x + 1}$

$so \: now \: then \: considering \\ \sqrt{x + 1} .cot \: Φ \\ \\ = \frac{PQ}{2} \times \frac{QR}{PQ} \\ \\ = \frac{QR}{2} \\ \\ \sqrt{x + 1}.cot \: Φ = \frac{x}{2}$

$now \: considering \: \\ \sqrt{x {}^{3} + x {}^{2} } .tan \: θ \\ \\ = \sqrt{x {}^{2}(x + 1) } .tan \: θ \\ \\ = x. \sqrt{x + 1} .tan \: θ \\ \\ = x. \frac{PQ}{2} \times \frac{QR}{PQ} \\ \\ = \frac{x.x}{2} \\ \\ \sqrt{x {}^{3} + x {}^{2} } .tan \:θ = \frac{x {}^{2} }{2}$

$similarl y\: then \\ cos \: θ = \frac{PQ}{PR} \\ \\ cos \: θ= \frac{2. \sqrt{x + 1} }{x + 2}$