IIT JEE , AIEEE Forum - Mathematics , Algebra

preetirocks

1. If in the trianglePQR, sinP, sinQ, sinR are in A.P. then
    a. the altitudes are in A.P.
    b.the altitudes are in H.P.
    c. the medians are in G.P.
    d. the medians are in H.P.

2. the distance between the line r - 2i - 2j + 3k +

(i - j+ 4k) and the plane r. (i + 5j + k) = 5. is

3. A variable line through the point (a,b) cuts the references at A and B respectively. The line through A and B are parallel to Y axis and X-axis respectively meets at P. Then the locus of P has the equation ????

4. Solution of the equation dy/dx + 1/x tan y = 1/x² tany siny

raulrag009

is tany in the denominator or numerator in the last question

preetirocks

tany is in the numerator only.

Decoder

1. & 3.
use sine rule..
sinp = sinq = sinr
----- ------ -----
p q r

add ..p,q eq's.. and equate it to middle one..
u get 2q = p+r ...........1)

now let alti's r l,m,n
thus lp = mq = nr....(area of triangle)
use... 1) to check it..

fro medians again use sine rule...(cut the upper angle(p) into two)..

3)

let the line is x/ m + y/n = 1..
now put (a,b)..as it passes through the line..

now m,n are intercepts...draw diagram u will find tht h=m & k=n..

thus locus is a/m + b/n = 1..
put m=h...and n=k..
u get a/x + b/y = 1...

preetirocks

Then wat will be the answer of first question

Sushmi

for 1 , altitudes are in H.P

See suppose the side opp to the angle P is p , angle Q is q and angle R is r

r sin P , p sin Q , q sin R are the altitudes

applying sine rule we have

= k sin Q sin R , k sin R sin P , k sin P sin R

= k sin P sin Q sin R / sin P , k sin P sin Q sin R / sin Q , k sin P sin Q sin R / sinR

so they are in H.P as the sine terms in A.P

Conjurer

1)

Area of triangle can be given as 1/2 Pm1 = 1/2 Qm2 = 1/2 R m3 where m1,m2,m3 are medians.

So m1 = 2delta/P
m2 = 2delta/Q
m3 = 2delta /R

But P Q R are in AP , hence m1 m2 m3 are in HP

elastiboysai

edited

elastiboysai

is tanysiny in d numerator or denom. in RHS?

Conjurer

Is the answer for 2) , 10/root27

@ elastiboy: sin@ =0 So angle between them is 90?

Sushmi

4.
Dividng numerator and denominator by tan y sin y ,

cot y cosec y dy/dx + 1/x cosec y = 1/x^2

Now put - cosec y = v

So dy/dx -1/x . v = 1/x^2

Now its in standard form

put 1/x as integrating factor

and we have 2x = sin y(1 - 2 c x^2)

preetirocks

ya...the answer for 2 is 10/root 27.

tany is in the numerator only

Decoder

is my answet 3) a/x + b/y = 1 ...correct ??

i hv solved it in my abv post..

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