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  • 2 answers

Rajat Arora 4 years, 8 months ago

In ΔABC, Let ∠ABC be θ Sin θ = (a2 - b2) / (a2 + b2) ⇒ AB = (a2 - b2) ⇒ AC = (a2 + b2) ⇒ BC = √[(a2 + b2)2 - (a2 - b2)2] [According to Pythagoras theorem] ⇒ BC = √(4a2b2) ⇒ BC = 2ab Cos θ = 2ab / (a2 + b2) Tan θ = (a2 - b2) / 2ab. Cosec θ, Sec θ and Cot θ are the reciprocals of Sin θ, Cos θ, Tan θ respectively.

Gaurav Seth 4 years, 8 months ago



In ΔABC, Let ∠ABC be θ

Sin θ = (a2 - b2) / (a2 + b2)

⇒ AB = (a2 - b2)

⇒ AC = (a2 + b2)

⇒ BC = √[(a2 + b2)2 - (a2 - b2)2]   [According to Pythagoras theorem]

⇒ BC = √(4a2b2)

⇒ BC = 2ab

Cos θ = 2ab / (a2 + b2)

Tan θ = (a2 - b2) / 2ab.

Cosec θ, Sec θ and Cot θ are the reciprocals of Sin θ, Cos θ, Tan θ respectively.

  • 1 answers

Sia ? 4 years, 4 months ago

The great Nature has intended us to earn our bread in the sweat of our brow. Every one, therefore, who idles away a single minute becomes to that extent a burden upon his neighbours, and to do so is to commit a breach of the very first lesson of ahimsa. Ahimsa is nothing if not a well-balance, exquisite consideration for one's neighbour, and an idle man is wanting in that elementary consideration. (YI, 11-4-1929, p. 144-15)

  • 1 answers

Ashok Kumar 4 years, 8 months ago

an=7000
  • 1 answers

Gaurav Seth 4 years, 8 months ago

 

AB is a tower. D and Care two points on the same side of a tower, BD = a and BC = b.

 

 

∠ADB and ∠ACB are the complementary angles.

 

If ∠ADB = x, then ∠ACB = 90 – x

 

In ∆ADB,

 

 

………… (1)

 

In ∆ABC,

 

 

…………....(2)

 

Multiplying (1) and (2),

 

 

 

(AB)2 = ab

 

AB = √ab

 

Height of tower = AB = √ab

 

Hence proved.

  • 2 answers

Pragati Pagare 4 years, 8 months ago

. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.  Solution: The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm) So, area of gold region = π r12 = π(10.5)2 = 346.5 cm2 Now, it is given that each of the other bands is 10.5 cm wide, So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm Thus, ∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2 = (π(21)2 − 346.5) cm2 = 1386 − 346.5 = 1039.5 cm2 Similarly, The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm For the area of nth region, A = Area of circle n – Area of circle (n-1) ∴ Area of blue region (n=3) = Area of third circle – Area of second circle = π(31.5)2 – 1386 cm2 = 3118.5 – 1386 cm2 = 1732.5 cm2 ∴ Area of black region (n=4) = Area of fourth circle – Area of third circle = π(42)2 – 1386 cm2 = 5544 – 3118.5 cm2 = 2425.5 cm2 ∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle = π(52.5)2 – 5544 cm2 = 8662.5 – 5544 cm2 = 3118.5 cm2

Aditya Kumar Jha 4 years, 8 months ago

Ok
  • 5 answers

Akanksha Bagiyal 4 years, 8 months ago

Ans I 1 As sin ^2 theta + cos ^2 theta = 1

Pragati Kumari 4 years, 8 months ago

1

Mayank Sharma 4 years, 8 months ago

1

Aditya Kumar Kumar 4 years, 8 months ago

1

Dhawal Khatri 4 years, 8 months ago

It is ncert identity and it's answer is 1
  • 2 answers

Aditya Kumar Kumar 4 years, 8 months ago

-x

Archi Jain 4 years, 8 months ago

  • 1 answers

Varun Agarwal 4 years, 8 months ago

For case study you can refer byjus or toppr or extra marks online (I don't promote any particular classes just for your help)
  • 1 answers

Gautam Sinha 4 years, 8 months ago

(1+cos¢\sin¢-1/sin¢) (1+sin¢/cos¢+1/cos¢) =sin²¢+cos²+2sincos¢-1÷cos¢sin¢ =1+2sincos¢-1÷sin¢cos¢ =2
  • 3 answers

Archi Jain 4 years, 8 months ago

Alpha + beta = - b/a Alpha × beta = c/a Alpha +beta +gama = -b/a Alpha ×beta× gama = c/a Alpha ×beta + beta × gama + gama × alpha = - d / a

Aashna Singh 4 years, 8 months ago

a+b = -b/a a×b = c/a a+b+y = -b/a ab+by+ya = c/a aby = -d/a a= alpha b= beta y= gamma

Piyush Piyush 4 years, 8 months ago

a+b=-b/a ab=c/a
  • 0 answers
  • 2 answers

Jatin Sharma 4 years, 9 months ago

Find the probality of getting a doublet in a throw of a pair of dice

Gaurav Seth 4 years, 9 months ago

Let the first number be x and the second number is 27 - x.
Therefore, their product = x (27 - x)
It is given that the product of these numbers is 182.
Therefore, x(27 - x) = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 13x - 14x + 182 = 0
⇒ x(x - 13) -14(x - 13) = 0
⇒ (x - 13)(x -14) = 0
Either x = -13 = 0 or x - 14 = 0
⇒ x = 13 or x = 14
If first number = 13, then
Other number = 27 - 13 = 14
If first number = 14, then
Other number = 27 - 14 = 13
Therefore, the numbers are 13 and 14.

  • 2 answers

Yogita Ingle 4 years, 9 months ago

7x² - 12x + 18

 

Find the sum of the roots:

Sum of the roots = α + β

α + β = - b/a

α + β = - (-12 ) / 7 = 12/7

 

Find the product of the roots:

Product of the roots = αβ

αβ = c/a

αβ = 18/7

Find the ratio:

α + β : αβ = 12/7 : 18/7

Multiply by 7:

α + β : αβ = 12 : 18

Divide by 6:

α + β : αβ = 2 : 3

Maxtern .Op 4 years, 9 months ago

14-12x+18 => 32=12x => x=32/12 => x=8/3
  • 1 answers

Yogita Ingle 4 years, 9 months ago

2x2 -7 x+3=0

2x2 -6x - 1x +3=0

2x ( x - 3) - 1( x- 3) = 0

(x - 3)(2x - 1) = 0

x - 3 = 0 or 2x - 1 = 0

x = 3 or x = 1/2

  • 2 answers

Neha Kumari 4 years, 9 months ago

Sec3A=Cosec(A-10) Sec3A=Sec(100-A) 3A=100-A 4A=100 A=25.............(ans)

Aanya Jain 4 years, 9 months ago

A is 25degree. sec(3A)=cosec (A-10) sec3A =sec(90-{A-10)} 3A=90+10-A 3A+A=100 4A=100 A=100/4 A=25
  • 2 answers

Aanya Jain 4 years, 9 months ago

a+2d=4. (1) a+8d=-8. (2) Subtract eq 1 and 2 a+2d=4 a+8d=-8 (-)(-). (+) --------------- -6d=12 d= -2 Replace in eq 1 a + 2(-2)=4 a-4=4 a=4+4 a=8 an=a+(n-1)d 0= 8+(n-1)-2 0-8 =(n-1)-2 -8/-2 =n-1 4=n-1 4+1 =n 5=n ---------
ನಂಗ್ ಗೊತ್ತಿಲ್ಲ??????
  • 1 answers

Yogita Ingle 4 years, 9 months ago

Given :  

Let ‘d’ be the diameter of a circle and ‘a’ be the side of a triangle.

Diameter of a circle =  side of equilateral triangle

d = a

Radius of a circle, r = d/2

 

Area of circle,A1 = πr²

Area of an equilateral ∆, A2 = √3/4 a²

Ratio of the Area of circle and Area of an equilateral ∆  :

A1 : A2  = πr² : √3/4× a²

A1 / A2  = πr² / √3/4 ×a²

A1 / A2  = π(d/2)² / √3/4× a²

A1 / A2  = π(a/2)² / √3/4 ×  a²

[d = a]

A1 / A2  = π(a²/4) / √3/4 × a²

A1 / A2  = πa²/4 ×  4 /√3a²

A1 / A2 = π/√3

A1 :  A2 = π : √3

Hence, the Ratio of the Areas of circle and equilateral ∆ is π : √3

  • 2 answers

Mayank Rauthan 4 years, 9 months ago

Let ABC be the triangle. The line l parallel to BC intersect AB at D and AC at E. To prove AD​/DB =AE​/EC Join BE,CD Draw EF⊥AB, DG⊥CA Since EF⊥AB, EF is the height of triangles ADE and DBE Area of △ADE=1/2 × base × height=1/2​AD×EF Area of △DBE= 1/2 ​×DB×EF areaofΔDBEareaofΔADE​=1/2​×AD×EF/ 1/2​×DB×EF​=AD/DB​ ........(1) Similarly, areaofΔDCEareaofΔADE​= 1/2 ×AE×DG/ 1/2 ×EC×DG ​=AE/EC​ ......(2) But ΔDBE and ΔDCE are the same base DE and between the same parallel straight line BC and DE. Area of ΔDBE= area of ΔDCE ....(3) From (1), (2) and (3), we have AD​/DB =AE​/EC Hence proved.

Yogita Ingle 4 years, 9 months ago

Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.

Let ABC be the triangle.
The line l parallel to BC intersect AB at D and AC at E.
To prove AD​/DB =AE​/EC
Join BE,CD

Draw EF⊥AB, DG⊥CA
Since EF⊥AB,
EF is the height of triangles ADE and DBE
Area of △ADE=1/2 × base × height=1/2​AD×EF
Area of △DBE= 1/2 ​×DB×EF
areaofΔDBEareaofΔADE​=1/2​×AD×EF/ 1/2​×DB×EF​=AD/DB​          ........(1)
Similarly,
areaofΔDCEareaofΔADE​= 1/2 ×AE×DG/ 1/2 ×EC×DG ​=AE/EC​            ......(2)

But ΔDBE and ΔDCE are the same base DE and between the same parallel straight line BC and DE.
Area of ΔDBE= area of ΔDCE         ....(3)
From (1), (2) and (3), we have

AD​/DB =AE​/EC

Hence proved.

  • 5 answers

Pratap Jadhav 4 years, 6 months ago

sand me the basic math Question

Shreeram Choudhary 4 years, 6 months ago

(1) 25 (2) 40 (3) 25 (4) 150 (5) 60

Shakthi Rock 4 years, 8 months ago

answer here ....

Divyansh Dotaniya 4 years, 8 months ago

d

Yogita Ingle 4 years, 1 month ago

Suresh is having a garden near Delhi. In the garden, there are different types of trees and flower plants. One day due to heavy rain and storm one of the trees got broken as shown in the figure.The height of the unbroken part is 15 m and the broken part of the tree has fallen at 20 m away from the base of the tree.
Using the Pythagoras answer the following questions:

1) What is the length of the broken part?

A) 15 m

B) 20 m

C) 25 m

D) 30 m

2) What was the height of the full tree?

A) 40 m

B) 50 m

C) 35 m

D) 30 m

3) In the formed right-angle triangle what is the length of the hypotenuse?

A) 15 m

B) 20 m

C) 25 m

D) 30m

4) What is the area of the formed right angle triangle?

A) 100 m2

B) 200 m2

C) 60 m2

D) 150 m2

5 )What is the perimeter of the formed triangle?

A) 60 m

B) 50 m

C) 45 m

D) 100 m

Answer:

1)C

2)A

3)C

4)D

5)A

  • 0 answers
  • 1 answers

Gaurav Seth 4 years, 9 months ago

Let the Tower be AC = 100m
Distance between cars is BD
BD=BC+CD
Let the BC be 'x' and CD be 'y'

In Traingle ACD
Cot45°= Base/Perpendicular
Cot45°=y/100
1=y/100
y=100m


In Traingle ACB
Cot30°=x/100
1.732=x/100
1.732×100=x
x=173.2m


Distance between cars = BC+CD
=x+y
=173.2+100
=273.2m

Or

  • 3 answers

Riya Rathore 4 years, 9 months ago

K=-15\4

Ashutosh Ranjan 4 years, 9 months ago

X ke place pe -1/2 daal do aur equate kro aa jayega

Gaurav Seth 4 years, 9 months ago

Putting x = (-1/2) in 3x2 + 2kx - 3 = 0,

or

Given : If  is a solution of the quadratic equation .

To find : The value of k ?

Solution :

To get the value of k substitute the value of x in the equation,

Equation 

Put ,

Therefore, The value of k is   or  

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