In the given figure, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM=1/2(∠Q−∠R) Class 09 math chapter triangle
in Triangle PQM : <Q + <QPM = 90 in Triangle RPM: <R + <RPM = 90 <Q - <R < <RPM - <QPM =(1/2 <P +<APM) - (1/>P -<APM) =2<APM search term in, triangle, pqr, prove, q, r, 2, apm in, given, fig, ,pa, bisector, ∠, qpr, pm, ⊥, qr
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Come, join us in this enriching educational experience, where curiosity is celebrated, creativity is nurtured, and learning is a lifelong passion. Step into the Student Think Zone at JSunil Tutorial, and let your imagination take flight. Happy thinking! In the given figure, each one of PA, QB and RC is perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that 1/x + 1/y= 1/z. by Jsunil sir in ∆QBC and ∆ PAC , ∠B = ∠A (90°) ∠C = ∠C (angles) ∆QBC ~ ∆PAC BQ/AP = BC/AC z/x = b/(a+b)-------(i) In and ∆QBA and ∆ ARC ∠B= ∠C ( 90°) ∠A = ∠A (common) in ∆ABQ ~∆ARC BO/RC = AB/AC z/y = a/(a+b)------(ii) From equation (i) and (ii), z(1/x +1/y) = 1 So, (1/x +1/y) = 1/z Key prove, 1x,1y,1z if, pa, qb, rc, perpendicular, ac, Class,10
In Fig OB is the perpendicular bisector of the line segment DE, FA ⊥ OB and FE intersects OB at the point C. Prove that 1/OA + 1/OB = 2/OC Solution by Jsunil
Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot if α and β be elevation of these station from the tower, Prove that its inclination θ to be Horizontal is given by Cot θ = (b cot α - acot β)/b-a
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