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Prince Kumar Prasad 6 years, 10 months ago

Solution is very lengthy so I can't post full solution.

Prince Kumar Prasad 6 years, 10 months ago

m=4.25 n=2.2
  • 1 answers

Simran ????? 6 years, 10 months ago

No. Becoz the HCF should be the factor of LCM of any given number.
  • 1 answers

Nikhar Sanadhya 6 years, 10 months ago

Sn=n/2(a+l) put the formula you get the value of n So the value of n = 16 So apply the formula l=a +(n-1) *d So you get the value of d
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Nishant Bansal 6 years, 10 months ago

12

Ayushi Singh 6 years, 10 months ago

Mean
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Nishant Bansal 6 years, 10 months ago

a^3b^3
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Jot Jattana 6 years, 10 months ago

21

Hitesh Rajpurohit 6 years, 10 months ago

12 / 21
  • 2 answers

Gaurav Seth 6 years, 10 months ago

Similar question:

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Give the justification of the construction.

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The tangents can be constructed on the given circles as follows.

Step 1

Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.

Step 2

Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents.

Justification

The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.

∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠ASB = 90°

⇒ BS ⊥ AS

Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

Anushka Jugran ? 6 years, 10 months ago

How can I draw here???
  • 2 answers

Nishant Bansal 6 years, 10 months ago

Bpt

Anushka Jugran ? 6 years, 10 months ago

Refer ncert
  • 2 answers

Surendrapal Singh Singh Gaur 6 years, 10 months ago

CSA of frustum is π(r1+r2)l Putting the values....answer will be 817.14 cm^2

Surendrapal Singh Singh Gaur 6 years, 10 months ago

It must b a frustum...
  • 2 answers

Ashish Chaudhary 6 years, 10 months ago

Sorry it is of objective (c)7/81(179+10 to the power -21)

Puja Sahoo? 6 years, 10 months ago

Kya answer 20.3 hai......
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Puja Sahoo? 6 years, 10 months ago

No

Puja Sahoo? 6 years, 10 months ago

AB'S oxford international school.......

Puja Sahoo? 6 years, 10 months ago

No, Because LCM>HCF and HCF of 2 numbers should divide the LCM completely. Here, 175 is not completely divisible by 15.
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Zeeshan Younis 6 years, 10 months ago

5/2√6
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Puja Sahoo? 6 years, 10 months ago

9,17,25.......
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Harshita Srivastava 6 years, 10 months ago

LHS= SinA/1+CosA+1+CosA/SinA(CosA/1+SinA+1+SinA/CosA) =Sin2+(1+CosA)2/SinA(1+CosA){Cos2A+(1+SinA)2/CosA(1+SinA)} =Sin2A(1+Cos2A+2CosA)/SinA+SinA×CosA(Cos2A+1+Sin2A+2SinA/CosA+SinA×CosA) =1+Sin2A+Cos2A+2CosA/SinA+SinA×CosA(1+Sin2A+Cos2A+2SinA/CosA+SinA×CosA) =2+2CosA/SinA(1+CosA){2+2SinA/CosA(1+SinA)} (Formula used : Sin2A+Cos2A = 1) =2(1+CosA)/SinA(1+CosA){2(1+SinA)/CosA(1+SinA} =2×1/SinA×1(2×1/CosA×1) =2/SinA(2/CosA) =2CosecA×2SecA (Formula used : 1- 1/SinA = CosecA 2- 1/CosA = SecA) =4CosecA×SecA So, L.H.S = R.H.S
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Divya Garg 6 years, 10 months ago

a+23d=2 (a+9d)....... a+23d=2a+18d....... 23d-18d=2a-a...... 5d=a....therefore a+71d=2 (a+33d).... a+71d=2a+66d..... 5d+71d=2 (5d)+66d..... 76d=10d+66d....76d=76d........ Hence proved....?
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Deepika Singh 5 years, 8 months ago

SecA=1/cosA TanA=sinA/cosA 1/cosA+sinA/cosA=P 1+sinA/cosA=P 1+sinA=PcosA SinA=Pcos-1
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Sia ? 6 years, 4 months ago

We know that, A + B + C = 180o 
{tex}\Rightarrow{/tex} A + B = 180o  - C
Dividing both sides with 2, we get
{tex}\Rightarrow{/tex} {tex}\frac{A + B}{2}{/tex} = 90o{tex}\frac{C}{2}{/tex}
Applying cosec on both sides, we get
{tex}\Rightarrow{/tex} cosec({tex}\frac{A + B}{2}{/tex}) = cosec(90o{tex}\frac{C}{2}{/tex})
{tex}\Rightarrow{/tex} cosec({tex}\frac{A + B}{2}{/tex}) = sec {tex}\frac{C}{2}{/tex}

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