Rakesh Dubey

BC is a side of a square, on the perpendicular bisector of BC, two points P, Q are taken, equidistant from the center of the square. BP, CQ are joined and cut in A.

In the Δ ABC, find the value of : tan A (tan B - tan C)² .

In a Δ ABC, ∠ ABC = 80°, ∠ ACB = 20°. D is a point on BC such that AB = DC

Find ∠ DAC. (Using Elementary Geometry and no Trigonometry please.)

Let α, β, γ > 0 and (α + β + γ) = π/2. If p = (tan α tan β) + 5, q = (tan β tan γ ) + 5

and r = (tan γ tan α) + 5 ,

find the maximum value of (√p + √q + √r ).

Solve for x and y ;

x + y + 5 + √(xy + 3x + 2y + 6) = 39 - - - - (i)

x² + y² + 7x + 8y + xy + 19 = 741 - - - - - -(ii)

Please follow up the link given below for the picture.

http://i854.photobucket.com/albums/ab104/RakeshDub...

Kedar sent a container of marbles by road from Mumbai to Pune. The container was 2m × 2m × 2.5m in size with height being the larger dimension. Marbles of 2 cm diameter were arranged to fill the entire base and then additional layers of marbles were arranged with each upper marble exactly on top of corresponding marble in the previous layer (as in the given figure ). The marbles were thus placed upto 2m height to complete the cubical structure. However, on reaching Pune, when opened the container, he found the height of structure was not 2m as promised by Kedar, but something else. Kedar defended saying that marbles may have readjusted due to jiggling. Can you find new height of this marble pile ?

Show that, If 1 < x < 2, then ►

[1 / √{ x + 2√( x - 1) }] + [1 / √{ x - 2√( x - 1) }] = 2 / ( 2 - x ) .

A 2 meter long object is fired vertically upwards from the mid-point of two locations A and B, 8 metres apart. The speed of the object after t seconds is given by ; (ds / dt) = (2t + 1) metres per second. Let α and β be the angles subtended by the object at A and B , respectively after one and two seconds. Find the value of cos (α − β)

In ∆ ABC, ∠ABC = 90°. D and E are points on AC such that | AD | = | AB |,and | CE | = | CB |. Let r be the in-radius of ∆ ABC and R be the circum-radius of ∆ BDE. Evaluate, R / r .

A person claimed that he saw a solar eclipse when the size of the solar disk was 26' and that of the lunar disk was 30'. He also claimed that at the the time of the maximum eclipse, distance between the centres of the two discs was 7'. Qualitatively show that he could not have observed a total eclipse. Find the percentage of the solar disc covered at the time of maximum eclipse. (Given: cos^-1 (1/26) ≈ 0.49π rad.) Thanks.

A ball moving around the circle, x² + y² – 2x – 4y – 20 = 0, in anti-clockwise direction leaves it tangentially at the point P(–2, –2). After getting reflected from a straight line it passes through the centre of the circle. Find the equation of the straight line if its perpendicular distance from P is 5/2.

A chord of a right triangle, which is parallel to the hypotenuse and passes through the incenter I, is divided by I into two segments of length m and n. Find the area of the triangle (in terms of m and n)

Here is the diagram;

http://i854.photobucket.com/albums/ab104/RakeshDub...

l , m and n are the lengths of the tangents to the incircle, drawn parallel to the sides BC , CA and AB respectively of a ∆ ABC.

Find the value of ; ( l / a) + ( m / b) + ( n / c)

{ where BC = a , CA = b and AB = c }

If a + b + c = 0 ; then evaluate :

[{(b - c)/a} + {(c - a)/b} + {( a - b )/c}][{a/(b - c)} + {b/(c - a)} + {c/(a - b)}]

If the expression;

{A cos (θ + α) + B sin (θ + β)} / {A' sin (θ + α) + B' cos (θ + β)}

retains the same value for all 'θ' , show that : --

AA' - BB' = (A'B - AB' ) sin (α - β) .

If ( cosec θ – sin θ ) = a^3 And ( secθ – cos θ ) = b^3 ,

then prove that : a²b² ( a² + b²) = 1.

As we know that there is no end of knowledge and a

Mathematics-question can be solved in different-different

ways/methods (shorter-smarter or lengthy-boring/tedious

or funny-interesting) yet some answerers show off, tend to

be over smart after giving their answers and they also comment

about other answerer's answers. They as well as think that if

they have attempted a question, no-one can now attempt

that question even if the question is still open to answer or

it shows,"left for 3 days" / "left for four days". After giving

their so-called nice and correct answer they think 'now

it is a copyright matter' so no-one can solve it better than that.

They don’t forget to distribute the certificates of 'stalkers, trolls

. ..etc, if you tried to solve that question after being solved by

them. They any how want to become a top answerer in two-

three weeks. I don't want to disclose the name of such

answerer as he is my contact-friend. What’s your opinion plz. ??

Should anyone not to answer such questions whose answers are already come at the screen even if the

the question is still open ? . . . . . .

Should anyone comment for other's answers rudely ?

Shouldn’t we report this to Y/A team, if anyone is doing these unfair ?

If the sum of first n terms of an A.P. is cn², find

the sum of squares of these n terms.

Evaluate: ∫ (x^27) * [(1 + x + x²)^6] * (6x² + 5x + 4)dx

In an isosceles triangle with base ' a ' and vertical angle 20°

and lateral side each ' b ' , prove that : a^3 + b^3 = 3ab^2.

Show that for any triangle with sides a, b and c;

3(ab + bc + ca) ≤ (a + b + c)² < 4(ab + bc + ca)