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TRIGONOMETRY KALPALATHA GHS SANTHIPURAM
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SOGADABALLA
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O P P O S I T E S I D E ADJESENTSIDE H Y P O T E N U S E A C B NAME OF THE SIEDES OF RIGHTANGLE TRIANGLE
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A OPPOSITESIDE ADJESENTSIDE H Y P O T E N U S E Sin θ= HYPOTENUSE OPPOSITESIDE ADJESENTSIDE θ
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A O P P O S I T E S I D E ADJESENTSIDE H Y P O T E N U S E Cos θ= HYPOTENUSE O P P O S I T E S I D E ADJESENTSIDE θ
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A OPPOSITESIDE ADJESENTSIDE H Y P O T E N U S E Cosec θ= HYPOTENUSE OPPOSITESIDE ADJESENTSIDE θ
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A O P P O S I T E S I D E ADJESENTSIDE H Y P O T E N U S E Sec θ= HYPOTENUSE O P P O S I T E S I D E ADJESENTSIDE θ
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O P P O S I T E S I D E H Y P O T E N U S E Tan θ= OPPOSITESIDE ADJESENTSIDE θ
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O P P O S I T E S I D E H Y P O T E N U S E Cot θ= OPPOSITESIDE ADJESENTSIDE θ
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Cotθ Secθ Cscθ Tanθ Cosθ Sinθ 90º60º45º30º0º 1 1/ √2 0 0 1/2√3/2 0 11/ √2 √3/2 1/2 1 1/ √3 √3 ∞ ∞∞ 22 √ 2 2/ √ 3 2/ √ 3 1 1 √ 2 ∞ 1 1/ √3 0
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ANGLE OF ELEVATION OBJECT HORIZONTAL OBSERVAR LINE SIGHT
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ANGLE OF DEPRESSION HORIZANTAL LIGHT OF SIGHT HORIZANTALOBJECT Angle of depression
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There are two temples, one on each bank of the river, just opposite to each Other.one of the temples A is 40 mts high.As observed from the top of this temple A, the angle of depression of the top and foot of the other temple B are and respectively. Find the width of the river and the height of the temple B approximately. 12°.30' 21°.48'
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12°.30' 12°.30' 21°.48' 21°.48' 90°
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E D C B A 21°.48' 12°.30' AB= river AC=BD=temples C=observer In right angle triangle ABC TanB=AC/AB Tan 21°.48' =40/AB 0.400 =40/AB AB =40/0.4 AB =100 mts In right angle triangle CED TanD=CE/ED Tan12°.30‘=CE/100 CE =0.2217×100=22.17 mts Height of the temple AC=AE+EC 40 =BD+22.17 ( BD=AE) BD =40-22.17=17.83 mts Height of the B temple =17.83 mts
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An aeroplane at an altitude of 2500 mts. Observes the angles of depression of opposite points on the two banks of river to be 41°.20' and 52°.10'. Find the width of the river in metres.
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41°.20' 52°.10' 41°.20' 52°.10' 2500mts
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A BC E AB=2500mts C and D are two bank river. CD is the width of the river Given ∟ACB= 41°.20‘ and ∟ ADB= 52°.10‘ From right angle triangle ACB tan 41°.20‘ =AB/CB 0.8795=2500/CB CB = 2500/0.8795 = 2842.5mts from right angle triangle ABD, ∟ADB = 52°.10‘ Tan 52°.10‘ = AB/BD 1.2876 = 2500/BD BD =2500/1.2876=1941.5mts CD=CB+BD =2842.5+1941.5=4784 mts The width of the river =4784 mts
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10th CLASS TEXT BOOK 9 th CLASS TEXT BOOK 8 th CLASS TEXT BOOK www.animatononline.com
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