MATHS Solution

Objective
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1b)sin x=2/3, y^2=3^2-2^2=5,y=5^(1/2),tan x=opp/adj=2/(5)^(1/2),cos x=5^(1/2)/3,tan x -cos x=5^(1/2)/15


13b) d surface area of d structure=surface area of mounted cone + surface area of the hemishere base= (22/7)7cmX(24cm^2 +7cm^2)^(1/2) +14cm=858cm^2

13)in <QRS  @=?, COS @=5cn/13cm=0.3846, @=(cos^-1)0.3846=67.38. in <PSR  [PR]=19cm [RS]=5cm [PS]=?,USING COSINE RULE [PS]^2=[PR]^2+[RS]^2-2[PR][RS]COS @, =19^2+5^2-2[19][5]COS67.38,  =312.926, [PS]=(312.926)^(1/2)=17.7cm.

11a)the perimeter of d wire mesh 13m, perimeter of the garden=2(3n-1)+ 2(2n+1), 2(3n-1)+2(2n+1)=135,6n-2+4n+2=135,6n-2+4n+2=135,10n=135,n=135/10=13.5



10a) contd?     $fx=3150,$f=40. mean =$fx/$f=3150/40=78.75. 

10b) S.D=(fd^2/f)^(1/2)=(2526.5/40)(1/2), =(63.1625)^(1/2)=7.95

10)table contd?. d^2:

280.5625,138.0625,45.5625,3.0625,10.5625,68.0625,175.5625,333.06
25.Fd^2=561.1250,414.1875,273.3750,33.6875,84.5000,476.4375,351.125,333.0625


10) table? class interval:60-64,65-69,70-74,75-79,80-84,85-89,90-94,95-99. 

F:2,3,6,11,8,7,2,1.   X= 62,67,72,77,82,87,92,97.

FX:124,201,432,847,656,609,184,9

7. d: -16.75,-11.75,-6.75,-1.75,3.25,8.25,13.25,18.25.


8c) @=40,TX=23.836m, HX=?, tan 40=TX/HX,HX=TX/tan40,=23.836/0.8391=28.41m.

8b)HT=?, using pythagora"s theorem,HT^2=TX^2+HX^2, =23.836^2 +20^2, HT=(968.154896)^(1/2),HT=31.1M. THE DISTANCE BTW H & D TOP OF THE TOWER IS 31.1m


8a)@=30,(HX)=20m,(PX)=?,using tan@=opp/adj,tan 30=PX/20m,(PX)=20mXtan30=11.546m. in triangle THX, @=50, TX=?, HX=20m, using tan@=opp/adj, tan 50=TX/20m,TX=20m X tan50,TX=23.836m. THUS PT=TX-PX,PT= 23.836-11.546=12.3m