# 1. in fig,a circle touches the side bc of triangle abc at p and touches ab and ac produced at q and r respectively if aq= 5cm then the perimeter of triangle abc=? 2. the value of 6+6+6+...................is 3.(xb-c)1bybc (xc-a)1byca(xa-b)1byab= 4.a sector is cut from a circle exceeds the diameter by 16.8 cm,then the length of arc is? 5.in an A.P 3rd term is 5 and the 7th term is 9 then 11th term is? 6.in a cyclic quadrilateral ABCD,cosA+cosB+cosC+cosD= 7.the value of log [(tan1)tan2)(tan3)...................(tan89)]is 8.median of xby 5,xby4,xby2,xby3is 8.if x>0,then thevalueof x is 9. the roots of the equation a(b-c)x2+b(c-a)x+c(a-b)=0 are

1
by vbreddymrt

2015-04-22T19:34:47+05:30

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See diagram.

AQ = AR = lengths of tangents from point A to the ex-circle of triangle ABC.
BQ = BP = lengths tangents from B onto the excircle.
CR = CP = lengths of tangents from C onto the excircle.
AQ = AB + BQ = AB + BP
AR = AQ = AC + CR = AC + CP
AQ + AR = 2 AQ = AB + BP + PC + CA  = perimeter of triangle ABC
So perimeter = 2 * 5 cm = 10 cm
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2)     6 + 6 + 6 + ....  n terms =  6 * n

3)    (xb - c) /bc  + (xc -a )/ca + (xa - b) / ab
= [ ax b - ac + x bc - ab + xac - bc ] / abc
=   (x - 1) ( ab + bc + ca) / abc
4)
case 1)   perimeter of the sector exceeds the diameter by 16.8 cm
(2 R + arc length) - 2 R = 16.8 cm      => arc length = 16.8 cm

case 2)  arc length exceeds
R Ф - 2 R = 16.8 cm.
there are two parameters R & Ф. we need more information to solve this.
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5)    T3 = 5 = a + 2 d
T7 = a + 8 d = 9
subtract equations,  6 d = 4      =>  d = 2/3
=>  a = 11/3
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6)    In a cyclic quadrilateral,  sum of opposite angles is 180 deg.
Cos A + Cos  C = 2 Cos (A+C)/2 Cos (A-C)/2    = 0
As (A+C)/2  = 90 deg
similarly  Cos B + Cos D = 0  as  (B+D)/2 = 90 deg.
=======================
7)  tan 1 tan2 tan 3  ...... tan 88 tan 89
= tan1  tan2 tan 3 .... tan 43  tan 44  tan 45  cot 44 cot 43 .... cot 3 cot 2  cot 1
= tan 1 cot 1  tan 2 cot2  tan3  cot3    .... tan43 cot43  tan44 cot44  * 1
= 1
Log  1 = 0
==============
8)
x / 5  , x/4  x/3  ,  x/2
median is the average of  x/4 and  x/3
so x/4 + x/3 = 2 * 8 = 16
7 x / 12 = 16
x = 12 * 16 / 7
======================
9)
D =  discriminant = b² (c-a)² - 4 ac (b-c) (a-b)
= b² (c² + a² - 2 a c) - 4 a c ( a b - b² - a c + b c)
= b² c² + a² b² - 2 a c b²  - 4 a² b c + 4 a c b² + 4 a² c² - 4 a b c²
= b² c² + a² b² + 2 a c b²  - 4 a² b c + 4 a² c² - 4 a b c²
= b² (a +c)² - 4 a c (a b - a c + b c)
= (a+c)² [ b² - 4 abc ] + 4 a² c²

x = [ - b (c - a) + - √ D ]  /  [ 2 a (b-c)  ]