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Question31.
A round balloon of radius r subtends an angle 'b' at the eye of the
observer while the angle of elevation of its centre is 'c'. Prove that
the height of the centre of the balloon is (r sin b cosec c/2) by Arshad Email add ares.ar...........@gmail.com answered on Feb 2, 2007 at 9:09 I.S.T.
(Figures cannot be drawn on our webpages) b is angle subtended subtended on eye of observer, so angle APB = b, where in figure PA Answer31. and PB act as tangents also join OA, OB and OP, O being the centre of circle.Triangle OAP is congruent to triangle OBP so angle APO = angle BPO = b/2.
angle OPC = c because it is angle of elevation.
sin c = h/OP and cosec b/2 = OP/r where h is height of centre of balloon
sin c ×cosec b/2 = h/OP × OP/r
= h/r
r sin c ×cosec b/2 = h
Question32. Prove that if Cos A + Sin A = root 2×Sin A , Then Sin A - Cos A = Root2×sin A
i think this question is wrong, please tell me if i am right. by Piyush Pattanayak piyush..........@gmail.com, answered on Feb 3, 2007 at 10:09 p.m. I.S.T.
Answer32. The correct question is Prove that if Cos A + Sin A = root 2 Sin A , Then Sin A - Cos A = Root2 cos A
Solution
Cos A = root 2 Sin A - Sin A
Cos A = (root 2 - 1 )Sin A
Sin A = cos A/(root 2 - 1 ) and then rationalising the denominator we get
sin A - cos A = root2 cosA
Question33. How to Factorise : 4 x^4 + y^4 (4 times x to the power of 4 plus y to the power of 4) by Saurabh , Email saurabh_bis........@yahoo.co.in, answered on Feb 8, 2007 at 8:52 p.m. I.S.T.
Answer33. 4 x^4 + y^4 can be solved by method of completing the squares
= [(2x)^2]^2 + (y^2)^2
= [(2x)^2]^2 + (y^2)^2 + 4x^2 y^2 - 4x^2 y^2 (adding and subtracting 2ab)
= [2x^2 + y^2]^2 - (2x^2y^2)^2 [ because a^2 + b^2 + 2ab = (a + b)^2]
= [2x^2 + y^2 - 2x^2y^2][2x^2 + y^2 + 2x^2y^2] [ a^2 - b^2 = (a - b)(a + b)]
Question34. (tanA+secA-1) / (tanA-secA+1) = (1 + sinA) / cosA by preeti bhanwala, Email em_pre010.........@yahoo.co.in, answered on 10 Feb 2007 at 1:22 p.m.
Answer34. L.H.S. = [tanA+secA-1] / [tanA-secA+1]
= [tanA+secA - (sec^2 A - tan^2 A)] / [tanA - secA + 1]
= [tanA+secA - (secA - tanA)(secA + tanA) / [tanA - secA + 1] [ a^2 - b^2 = (a - b)(a + b)]
= (tanA+secA) (tanA - secA + 1) / [tanA - secA + 1]
= tanA+secA
= (1 + sinA) / cosA
= R.H.S.
Question35. Can is substitute the value for tan 45* in a question which specifies that "we should not use trigonometric table" ? by T.Santhosh Kumar santhoshney.......@gmail.com, answered on Feb 11, 2007 at 1:20 p.m.
Answer35. Yes, we put the value of tan 45* in these questions. Replace tan 45* by 1, its value.
(Figures cannot be drawn on our webpages) b is angle subtended subtended on eye of observer, so angle APB = b, where in figure PA Answer31. and PB act as tangents also join OA, OB and OP, O being the centre of circle.Triangle OAP is congruent to triangle OBP so angle APO = angle BPO = b/2.
angle OPC = c because it is angle of elevation.
sin c = h/OP and cosec b/2 = OP/r where h is height of centre of balloon
sin c ×cosec b/2 = h/OP × OP/r
= h/r
r sin c ×cosec b/2 = h
Question32. Prove that if Cos A + Sin A = root 2×Sin A , Then Sin A - Cos A = Root2×sin A
i think this question is wrong, please tell me if i am right. by Piyush Pattanayak piyush..........@gmail.com, answered on Feb 3, 2007 at 10:09 p.m. I.S.T.
Answer32. The correct question is Prove that if Cos A + Sin A = root 2 Sin A , Then Sin A - Cos A = Root2 cos A
Solution
Cos A = root 2 Sin A - Sin A
Cos A = (root 2 - 1 )Sin A
Sin A = cos A/(root 2 - 1 ) and then rationalising the denominator we get
sin A - cos A = root2 cosA
Question33. How to Factorise : 4 x^4 + y^4 (4 times x to the power of 4 plus y to the power of 4) by Saurabh , Email saurabh_bis........@yahoo.co.in, answered on Feb 8, 2007 at 8:52 p.m. I.S.T.
Answer33. 4 x^4 + y^4 can be solved by method of completing the squares
= [(2x)^2]^2 + (y^2)^2
= [(2x)^2]^2 + (y^2)^2 + 4x^2 y^2 - 4x^2 y^2 (adding and subtracting 2ab)
= [2x^2 + y^2]^2 - (2x^2y^2)^2 [ because a^2 + b^2 + 2ab = (a + b)^2]
= [2x^2 + y^2 - 2x^2y^2][2x^2 + y^2 + 2x^2y^2] [ a^2 - b^2 = (a - b)(a + b)]
Question34. (tanA+secA-1) / (tanA-secA+1) = (1 + sinA) / cosA by preeti bhanwala, Email em_pre010.........@yahoo.co.in, answered on 10 Feb 2007 at 1:22 p.m.
Answer34. L.H.S. = [tanA+secA-1] / [tanA-secA+1]
= [tanA+secA - (sec^2 A - tan^2 A)] / [tanA - secA + 1]
= [tanA+secA - (secA - tanA)(secA + tanA) / [tanA - secA + 1] [ a^2 - b^2 = (a - b)(a + b)]
= (tanA+secA) (tanA - secA + 1) / [tanA - secA + 1]
= tanA+secA
= (1 + sinA) / cosA
= R.H.S.
Question35. Can is substitute the value for tan 45* in a question which specifies that "we should not use trigonometric table" ? by T.Santhosh Kumar santhoshney.......@gmail.com, answered on Feb 11, 2007 at 1:20 p.m.
Answer35. Yes, we put the value of tan 45* in these questions. Replace tan 45* by 1, its value.
Question36. For Income Tax, in which case the surcharge will be calculated -- when
the gross income exceeds Rs.10,00,000 or when the taxable income
exceeds Rs.10,00,00 ?? And then after that on what will be the edu cess
calculated -- edu cess will be 2% of tax (excluding surcharge) or 2% of
total tax (including surcharge) ?? Plzz help. by Pinaakee Bhardwaj, Email pinaak.....@yahoo.co.in, answered on Feb 12, 2007 at 11 p.m. I.S.T.
Answer 36.See table below which clears your doubt, if you still have any problem fill in the form on our website once again.
Question37. A triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that angle BAL = angle ACB. by shahid akhtar, Email akhtar........@yahoo.co.in, answered on Feb 13, 2007 at 1:12 p.m. I.S.T.
Answer37. In triangle BAC, angle B + angle C + angle BAC = 180 ........(1)
In triangle BLA, angle B + angle BLA + angle BAL = 180 ........ (2)
From (1) and (2) angle B + angle C + angle BAC = angle B + angle BLA + angle BAL
angle C + angle BAC = angle BLA + angle BAL
angle ACB = angle BAL (because angle BLA = angle BAC = 90)
Question 38. Sir, is there any changes in the income tax table for 2007 class 10 mathematics examination ? by suresh babu, Email sureshbabuko......@yahoo.co.in, answered on Feb 14, 2007 at 6:31 p.m.
Answer38. There is no change.
Question 39. simplify ___x___ + __y___ + ___z___
(x-y)(x-z) (y-z) ( y - x) ( z - x) ( z - y) by khushboo sharma Email khush_s_........@yahoo.co.in, answered on Feb 15, 2007 at 8:24 I.S.T.
Answer 39. ___x___ - __y___ + ___z___
(x-y) (x-z) (y-z) ( - y + x) ( - z + x) (- z + y) [by taking -1 common from (y-x) (z-x) (z-y)]
= [x(y - z) - y(x - z) + z(x - y)] / (x - y)(y - z)(x - z)
= 0 / (x - y)(y - z)(x - z)
= 0
Question 40.. S is a point in the interior of a triangle ABC.Prove that AB+AC > SB+SC by Akshay Shreedhar class 1X, Email akshayshree.........@gmail.com, answered on Feb 19, 2007 at 7:45 p.m.
Answer 40.
Construction - Produce BS to intersect AC at T
Proof - In triangle ABT, AB + AT > BT ........(i)
In triangle STC, ST + TC > SC ........(ii)
Adding (i) and (ii)
AB + AT + ST + TC > BT + SC
AB + AT + ST + TC > SB + ST + SC (BT = BS + ST from figure)
AB + AT + TC > SB + SC ( cancelling ST)
AB + AC > SB + SC (AT + TC = AC from figure)
Answer 36.See table below which clears your doubt, if you still have any problem fill in the form on our website once again.
|
Taxable Income |
Rate of Tax |
Surcharge |
Educational cess |
|
Upto Rs.1,35,000 |
Nil |
Nil |
Nil |
|
Rs.1,35,001 to
Rs.1,50,000 |
10% of
Total Income exceeding Rs.1,35,000 |
Nil |
2% of income tax |
|
Rs. 1,50,001
to Rs.2,50,000 |
Rs.1500 + 20% of total income exceeding Rs.1,50,000 |
Nil |
2% of income tax |
|
Rs.2,50,001
to Rs.10,00,00 |
Rs.21,500 + 30% of total |
Nil |
2% of income tax |
|
Above Rs.10,00,000 |
Rs.246500 + 30%
of total Income exceedingRs.10,00,000 |
10% of income tax |
2% of income tax and surcharge |
Question37. A triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that angle BAL = angle ACB. by shahid akhtar, Email akhtar........@yahoo.co.in, answered on Feb 13, 2007 at 1:12 p.m. I.S.T.
Answer37. In triangle BAC, angle B + angle C + angle BAC = 180 ........(1)
In triangle BLA, angle B + angle BLA + angle BAL = 180 ........ (2)
From (1) and (2) angle B + angle C + angle BAC = angle B + angle BLA + angle BAL
angle C + angle BAC = angle BLA + angle BAL
angle ACB = angle BAL (because angle BLA = angle BAC = 90)
Question 38. Sir, is there any changes in the income tax table for 2007 class 10 mathematics examination ? by suresh babu, Email sureshbabuko......@yahoo.co.in, answered on Feb 14, 2007 at 6:31 p.m.
Answer38. There is no change.
Question 39. simplify ___x___ + __y___ + ___z___
(x-y)(x-z) (y-z) ( y - x) ( z - x) ( z - y) by khushboo sharma Email khush_s_........@yahoo.co.in, answered on Feb 15, 2007 at 8:24 I.S.T.
Answer 39. ___x___ - __y___ + ___z___
(x-y) (x-z) (y-z) ( - y + x) ( - z + x) (- z + y) [by taking -1 common from (y-x) (z-x) (z-y)]
= [x(y - z) - y(x - z) + z(x - y)] / (x - y)(y - z)(x - z)
= 0 / (x - y)(y - z)(x - z)
= 0
Question 40.. S is a point in the interior of a triangle ABC.Prove that AB+AC > SB+SC by Akshay Shreedhar class 1X, Email akshayshree.........@gmail.com, answered on Feb 19, 2007 at 7:45 p.m.
Answer 40.
Construction - Produce BS to intersect AC at T
Proof - In triangle ABT, AB + AT > BT ........(i)
In triangle STC, ST + TC > SC ........(ii)
Adding (i) and (ii)
AB + AT + ST + TC > BT + SC
AB + AT + ST + TC > SB + ST + SC (BT = BS + ST from figure)
AB + AT + TC > SB + SC ( cancelling ST)
AB + AC > SB + SC (AT + TC = AC from figure)
Question 41. A peacock is sitting on a tree nine metres high. A snake at a distance of twenty Seven metres from pillar is coming to a pole at the base of the pillar. Seeing the Snake the peacock pounces upon it. If their speeds are equal find the distance from the hole at which the snake caught. by kartarlal class X, E mail kartar.l....@rediff.com, answered on Feb 20, 2007 at 8:32 p.m.
Answer41. Let AB represent tree, C is position of snake. It is moving along CB. Let D be the position where snake is caught by the peacock.
Let CD = x metres. Then AD = x metres where AD is distance covered by the peacock. BD = (27 - x) metres
Now in right triangle ABD, AB^2 + BD^2 = AD^2
9^2 + (27 - x)^2 = x^2
x^2 - (27 - x)^2 = 81
(x - 27 + x)(x + 27 - x) = 81
(2x - 27)27 = 81
(2x - 27) = 3
2x = 30
x = 15
Required distance = 27 - 15 = 12 metres
Question 42.A triangle ABC is right angled at B. AB=a BC=b and AC=c.The perpendicular from B on AC is of length 'p'.Prove that
1/p^2=1/a^2 + 1/b^2 by Mustafa, E mail mustafa.k......@gmail.com, answered at Feb 21, 2007 at 5:41 p.m.
Answer42. ar(triangle ABC) = ar(right triangle ABC)
1/2 (cp) = 1/2(ab) [area of triangle = 1/2 base×height]
cp = ab
c = ab/p .........(i)
In right triangle ABC, c^2 = b^2 + a^2
(ab)^2/p^2 = b^2 + a^2 using(i)
1/p^2=1/a^2 + 1/b^2 dividing both sides (ab)^2
Question 43. sir, can you please let me know, on which website can i see the new cbse syllabus for 2007-2008, for x standard. i have gone through all the possible sites but dont see the new syllabus anywhere. kindly help. by nayana anil, Email manjula_a.........@yahoo.com, answered on Feb 22, 2007 at 8:09 p.m.
Answer43. You can have it on our site at http://www.cbsemath.com/boarddownloads.htm which is given as cbse syllabus 2006 - 2008
Question 44. ABC and ADC are two right triangles with common hypotenuse CD. Prove that angle CAD = angle CBD. by sandeep tiwari, Email sibu_789632.......@rediff.com , answered on Feb 23, 2007 at 6:19 p.m.
Answer 44. There can be 2 cases (i) when triangles are on opposite sides of CD (ii)when triangles are on same sides of CD
case i
In quadrilateral ABCD, angle B + angle D = 90 + 90 = 180. Therefore quadrilateral ABCD is cyclic.
So points A, B, C and D are concyclic. angle CAD = angle CBD (angles in same segment)
case ii
angle B and angle D are equal and on same side of AC. Therefore points A, B, C and D are concyclic.
angle CAD = angle CBD (angles in same segment)
Question 45. sir, I want to publish some sample papers in maths for class X cbse students in my name. Please tell me how I can upload my files or how can i do above, by Santanu Kumar Mishra, Email santanuchemist....@gmail.com, answered on Feb 25, 2007 at 11:57 a.m.
Answer 45. You can send them to contact@cbsemath.info as attachments, we will publish them on cbsemathspapers.com with your name. You can send papers from class V1 to X11.
Answer41. Let AB represent tree, C is position of snake. It is moving along CB. Let D be the position where snake is caught by the peacock.
Let CD = x metres. Then AD = x metres where AD is distance covered by the peacock. BD = (27 - x) metres
Now in right triangle ABD, AB^2 + BD^2 = AD^2
9^2 + (27 - x)^2 = x^2
x^2 - (27 - x)^2 = 81
(x - 27 + x)(x + 27 - x) = 81
(2x - 27)27 = 81
(2x - 27) = 3
2x = 30
x = 15
Required distance = 27 - 15 = 12 metres
Question 42.A triangle ABC is right angled at B. AB=a BC=b and AC=c.The perpendicular from B on AC is of length 'p'.Prove that
1/p^2=1/a^2 + 1/b^2 by Mustafa, E mail mustafa.k......@gmail.com, answered at Feb 21, 2007 at 5:41 p.m.
Answer42. ar(triangle ABC) = ar(right triangle ABC)
1/2 (cp) = 1/2(ab) [area of triangle = 1/2 base×height]
cp = ab
c = ab/p .........(i)
In right triangle ABC, c^2 = b^2 + a^2
(ab)^2/p^2 = b^2 + a^2 using(i)
1/p^2=1/a^2 + 1/b^2 dividing both sides (ab)^2
Question 43. sir, can you please let me know, on which website can i see the new cbse syllabus for 2007-2008, for x standard. i have gone through all the possible sites but dont see the new syllabus anywhere. kindly help. by nayana anil, Email manjula_a.........@yahoo.com, answered on Feb 22, 2007 at 8:09 p.m.
Answer43. You can have it on our site at http://www.cbsemath.com/boarddownloads.htm which is given as cbse syllabus 2006 - 2008
Question 44. ABC and ADC are two right triangles with common hypotenuse CD. Prove that angle CAD = angle CBD. by sandeep tiwari, Email sibu_789632.......@rediff.com , answered on Feb 23, 2007 at 6:19 p.m.
Answer 44. There can be 2 cases (i) when triangles are on opposite sides of CD (ii)when triangles are on same sides of CD
case i
In quadrilateral ABCD, angle B + angle D = 90 + 90 = 180. Therefore quadrilateral ABCD is cyclic.
So points A, B, C and D are concyclic. angle CAD = angle CBD (angles in same segment)
case ii
angle B and angle D are equal and on same side of AC. Therefore points A, B, C and D are concyclic.
angle CAD = angle CBD (angles in same segment)
Question 45. sir, I want to publish some sample papers in maths for class X cbse students in my name. Please tell me how I can upload my files or how can i do above, by Santanu Kumar Mishra, Email santanuchemist....@gmail.com, answered on Feb 25, 2007 at 11:57 a.m.
Answer 45. You can send them to contact@cbsemath.info as attachments, we will publish them on cbsemathspapers.com with your name. You can send papers from class V1 to X11.
For question numbers 46 onward visit page 4