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Asked on 15 Apr CBSE/Class 10/Mathematics

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Lesson Posted on 13 Apr CBSE/Class 10/Mathematics

Chemistry Concept

Manibhushan Kumar

I have 15 years of experience in Maths and Stats.I have started teaching 1 year after completing my graduate...

Chemistry Concept. Overview; Acid-Base Reactions; Buffers; Catalysis; ChemicalEquilibrium; Empirical and Molecular Formulas; Electrochemistry; Electrolysis; Electron Quantum
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Lesson Posted on 13 Apr CBSE/Class 10/Mathematics

The log function

Sudhansu Bhushan R.

I am teaching Physics, Chemistry and Mathematics for Class - IX and class X in respect of I.C.S.E.,...

The log function Background: Any positive number, y, can be written as 10 raised to some power, x. We can write this relationship in equation form: y = 10 x For example it is obvious that 1000 can be written as 10 3. It is not so obvious that 16 can be written as 10 1.2. How do we know that this is... read more

The log function

Background: Any positive number, y, can be written as 10 raised to some power, x. We can write this relationship in equation form:

y = 10 x

For example it is obvious that 1000 can be written as 10 3. It is not so obvious that 16 can be written as 10 1.2. How do we know that this is the correct power of 10? Because we get it from the graph shown below.

To make this graph we made a table of a few obvious values of  y = 10 x  as shown below, left. Then we plotted the values in the graph (they are the red dots) and drew a smooth curve through them. Then we observed that the curve went through y = 16 and x = 1.2 (the black dot). This means that 16 = 10 1.2.


   



We next define a function called the logarithm that takes a number like 16 as input, calculates that it can be written as 10 1.2, and returns the exponent 1.2 as its output value. Here is the formal definition.


Definition: log(x) is defined as the function that takes any positive number x as input and returns the exponent to which the base 10 must be raised to obtain x



Example 1:   Evaluate log ( 10 5.7 ). In this example the argument of the log function is already expressed as 10 raised to an exponent, so the log function simply returns the exponent:

log ( 10 5.7 ) = 5.7


Example 2:   Evaluate log ( 1000 ). The argument is a number that is easily expressed as 10 raised to an exponent. We do this and the log function then returns the exponent:

log ( 1000 ) = log ( 10 3 ) = 3


Example 3:   Evaluate log ( 16 ). The argument is a number which we don't know how to express as 10 raised to an exponent (unless we remember the above discussion which said that 16 = 10 1.2 ). Therefore we use a calculator or the Algebra Coach to evaluate it:

log ( 16 ) = 1.2


Example 4:   log ( x ). The argument is an expression. Until we can evaluate that expression we must leave this logarithm as is.



Graph: The blue curve shown to the right is the graph of the log function. Notice that for any positive x it returns a single value. For any negative x it is undefined. If you compare this graph of the log function to thegraph of y = 10 x then you see that one can be gotten from the other by interchanging the x and y axes.

For comparison we have also shown the graph of the ln function (the natural log function) in red. The ln graph has the same shape as the log graph but is 2.3 times as tall. The ln function is described below.

An important feature of the log function is that it increases very slowly as x becomes very large. It describes nicely how the human ear percieves loud sounds and the way the human eye percieves bright lights.



Domain and range: The domain of the log function is all positive real numbers and the range is all real numbers.

The log function can be extended to the complex numbers, in which case the domain is all complex numbers except zero. The logarithm of zero is always undefined. 



Some special values of the log function




Solving the Equation 10 x = y for x by using the log function

Suppose that x is unknown but that 10 x equals a known value y. Then finding x requires solving the following equation for x:

10 x = y

The solution is:

x = log (y)

This is because finding log (ymeans expressing y as 10 to an exponent, and then returning that exponent. But the original equation says that that exponent is x. Note that if y is negative then there is no real solution. However there is a complex solution. Furthermore if y = 0 then there is no solution at all. 



log (x) and 10 x are inverse functions

Consider the 10 x function which takes x and returns 10 x, like this:

The log function is defined to do exactly the opposite, namely:

Therefore these are inverse functions.

Note the following:

  • Because the 10 x function is the inverse of the log function it is sometimes called the antilog function. 
  • We saw above that the solution of 10 x = y  is  x = log (y). We should look at these two equations as expressing the same relationship between x and ybut from different points of view. The first equation is the relationship solved for y and the second one is the relationship solved for x. (An analogy is that the statement “Tom is Jane’s brother” is equivalent to the statement that “Jane is Tom’s sister”.) 
  • In the previous bullet we saw that the two equations, 10 x = y  and  x = log (y), said the same thing. If we replace x in the first equation by the x of the second equation we get this identity:

10 log( y) = y

and if we replace y in the second equation by the y of the first equation we get this identity:

x = log (10 x )

These identities are useful for showing how the log and antilog functions cancel each other.

  • If you compare the graph of y = log (x) to the graph of y = 10 x then you see that one can be gotten from the other by interchanging the x and y axes. This always happens with inverse functions.

 


How to use the log function in the Algebra Coach

  • Type log(x) into the textbox, where x is the argument. The argument must be enclosed in brackets. 
  • Set the relevant options:
    • Set the exact / floating point option. In floating point mode the log of any number is evaluated. In exact mode the log of an integer is not evaluated because doing so would result in an approximate number.
    • Turn on complex numbers if you want to be able to evaluate the log of a negative or complex number.
  • Click the Simplify button. 

 


Algorithm for the log function

Click here to see the algorithm that computers use to evaluate the log function. 



The ln function

Background: You might find it useful to read the previous section on the log function before reading this section. Recall that the log function takes a number like 16 as input, calculates that it can be written as 10 1.2, and then returns the exponent 1.2 as its output value. But why use base 10? After all, probably the only reason that the number 10 is important to humans is that they have 10 fingers with which they first learned to count. Maybe on some other planet they use base 8!

In fact probably the most important number in all of mathematics (click here to see why) is the number 2.7182818284590…, which we give the name e, in honor of Leonard Euler, who first discovered it. It will be important to be able to take any positive number, y, and express it as e raised to some power, x. We can write this relationship in equation form:

y = e x

For example 5 can be written as e 1.6 (the exponent is approximate). How do we know that this is the correct power of e? Because we get it from the graph shown below.

To make this graph we made a table of a few obvious values of  y = e x  as shown below, left. Then we plotted the values in the graph (they are the red dots) and drew a smooth curve through them. Then we observed that the curve went through y = 5 and x = 1.6 (the black dot). This means that 5 = e 1.6


   


If you compare this graph to that of y = 10 x you see that both have the same so-called exponential growth shape but that this graph grows more slowly.


We next define a function called the natural logarithm that takes a number like 5 as input, calculates that it can be written as e 1.6, and returns the exponent 1.6 as its output value. Here is the formal definition.


Definition: ln(x) is defined as the function that takes any positive number x as input and returns the exponent to which the base e must be raised to obtain x. (e denotes the number 2.7182818284590…) 


From now on, to avoid confusion, we will call ln(x) the natural logarithm function and log(x) the base 10 logarithm function.


Example 1:   Evaluate ln ( e 4.7 ). In this example the argument of the ln function is already expressed as e raised to an exponent, so the ln function simply returns the exponent:

ln ( e 4.7 ) = 4.7


Example 2:   Evaluate ln ( 8.6 ). The argument is a number which we don't know how to express as e raised to an exponent . Therefore we use a calculator or the Algebra Coach to evaluate it:

ln ( 8.6 ) = 2.15


Example 3:   ln ( x ). The argument is an expression. Until we can evaluate that expression we must leave this natural logarithm as is. 


Example 4:   Evaluate ln ( e ). Express the argument as e raised to the exponent 1 and return the exponent:

ln ( e ) = ln ( e 1 ) = 1


Example 5:   Evaluate ln ( 1 ). Express the argument as e raised to the exponent 0 and return the exponent:

ln ( 1 ) = ln ( e 0 ) = 0




Graph: The red curve shown to the right is the graph of the ln function. Notice that for any positive x it returns a single value. For any negative x it is undefined. If you compare this graph of the ln function to thegraph of y = e x then you see that one can be gotten from the other by interchanging the x and y axes.

For comparison we have also shown the graph of the base 10 log function in blue. The log graph has the same shape as the ln graph but is only 43% as tall. 



Domain and range: The domain of the ln function is all positive real numbers and the range is all real numbers.

The ln function can be extended to the complex numbers, in which case the domain is all complex numbers except zero. The natural logarithm of zero is always undefined. 



Solving the Equation e x = y for x by using the ln function

Suppose that x is unknown but that e x equals a known value y. Then finding x requires solving this equation for x:

e x = y

The solution is:

x = ln (y)

because finding ln (ymeans expressing y as e to an exponent, and then returning that exponent. But the original equation says that that exponent is x. Note that if y is negative then there is no real solution. However there is a complex solution. Furthermore if y = 0 then there is no solution at all. 



ln (x) and e x are inverse functions

Consider the e x function which takes x and returns e x, like this:

The ln function is defined to do exactly the opposite, namely:

Therefore these are inverse functions.

Note the following:

  • We saw above that the solution of e x = y  is  x = ln (y). We should look at these two equations as expressing the same relationship between x and y but from different points of view. The first equation is the relationship solved for y and the second one is the relationship solved for x. (An analogy is that the statement “Tom is Jane’s brother” is equivalent to the statement that “Jane is Tom’s sister”.) 
  • In the previous bullet we saw that the two equations, e x = y  and  x = ln (y), said the same thing. If we replace x in the first equation by the x of the second equation we get this identity:

e ln (y) = y

and if we replace y in the second equation by the y of the first equation we get this identity:

x = ln (e x)

These identities are useful for showing how the ln and e x functions cancel each other.

  • If you compare the graph of y = ln (x) to the graph of y = e x then you see that one can be gotten from the other by interchanging the x and y axes. This always happens with inverse functions.

 


How to use the ln function in the Algebra Coach

  • Type ln(x) into the textbox, where x is the argument. The argument must be enclosed in brackets. 
  • Set the relevant options:
    • Set the exact / floating point option. In floating point mode the ln of any number is evaluated. In exact mode the ln of an integer is not evaluated because to do so would result in an approximate number.
    • Set the e does / does not represent 2.718… option. (Set this to does represent so that for example ln(e 3 ) simplifies to 3.)
    • Turn on complex numbers if you want to be able to evaluate the ln of a negative or complex number.
  • Click the Simplify button. 

 


Algorithm for the ln function

Click here to see the algorithm that computers use to evaluate the natural logarithm function. 



The exp or e x function

Background: You might find it useful to read the previous section on the ln function before reading this section. There we saw that it is possible to use the number e (which is approximately 2.7182818284590…) as a base and to raise it to any power, x, and produce any positive number y. We can write this relationship in equation form:

y = e x

Here is a graph of y = e x (the blue curve). For comparison we also show graphs of y = 2 x and y = 4 x. Because the number e is between 2 and 4 the curve y = e x lies between the curves y = 2 x and y = 4 x.



All three of these curves are called exponential functions because the independent variable x is in the exponent. All three have the property that the higher up the curve you go the steeper they get. However y = e x has the special property that at every point along the curve the slope equals the height. This property concerning slopes makes it a very important function in calculus.


Definition: We define a function called the exp function that takes an argument x and returns the value of e raised to the power x:


You can think of exp(x) as just an alternative (functional) notation for the expression e x. So of course the functional form exp(x) has all the properties that the exponential form e x has. The Algebra Coach has an option that allows you to use one form or the other. Here is a table comparing the “look” of the various properties in the two forms:

Functional form

Exponential form

 
   

info on this property

   

info on this property

   

info on this property

   

info on this property

   

info on this property

   

info on this property

   

info on this property

   

info on this property





Graph: The blue curve is the graph of  y = e x  (i.e. of the exp function). It has the property that its slope equals its height everywhere. The dotted red lines show the slope of the curve at various points along the curve. Notice that the slope is 5 when the height is 5, and so on.

If you compare this graph of the exp function to the graph of the ln function then you see that one can be gotten from the other by interchanging the x and y axes.

Domain and range: The domain of the exp function is all real numbers and the range is all positive real numbers. 

The exp function can be extended to the complex numbers, in which case the domain and the range is all complex numbers. 


Solving the Equation ln (x) = y for x by using the e x or exp function

Suppose that x is unknown but that ln (x) equals a known value y. Then finding x requires solving this equation for x:

ln (x) = y.

The solution is:

x = e y,    or    x = exp (y).

This was explained in the previous section on the ln function



ln (x) and e x are inverse functions

This was explained in the previous section on the ln function.


How to use the exp function in the Algebra Coach

  • Type exp(x) into the textbox, where x is the argument. The argument must be enclosed in brackets. 
  • Set the relevant options:
    • Set the exponential function option to this: 
    • Set the exact / floating point option. In floating point mode the exp of any number is evaluated. In exact mode the exp of an integer is not evaluated because to do so would result in an approximate number.
    • Turn on complex numbers if you want to be able to evaluate the exp of a complex number.
  • Click the Simplify button. 

Note: You never need to use the functional notation exp (x) in the Algebra Coach. You can always use the notation e x instead. In fact the default setting for the exponential function option is exp (x) → e x


Algorithm for the e x or exp function

Click here to see the algorithm that computers use to evaluate e x or exp (x).

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Lesson Posted on 03 Apr CBSE/Class 10/Mathematics

Trigonometry

Lino P Thomas

I can teach each and every topics very easily and make it stronger for the students. I have an experience...

Hello Students, Here I am for you to provide some simple and basic lessons on how to deal with trigonometry Hopefully, You all would be aware of Sine, Cosine, Tangent Functions and their reciprocals. We can easily understand these functions through drawing a triangle ABC and marking an angle (say... read more

Hello Students,

Here I am for you to provide some simple and basic lessons on how to deal with trigonometry

Hopefully, You all would be aware of Sine, Cosine, Tangent Functions and their reciprocals.

We can easily understand these functions through drawing a triangle ABC and marking an angle (say at A) inside it.

Then,

Sin A = Opposite side / Hypotenuse

Cos A = Adjacent Side / Hypotenuse

Tan A = Opposite side / Adjacent side

Cosec A, Sec A, Cot A would be their reciprocals respectively.

 

Hope you guys are clear in this session.

Stay tuned for more .. If you like this, please provide your valuable feedback.

Thanks!!

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Answered on 02 Apr CBSE/Class 10/Mathematics

How to explain trigonometry application

Ayush Ranjan

Facilitator

In construction we need trignometry to calculate the following: Measuring fields, lots and areas; Making walls parallel and perpendicular; Installing ceramic tiles; Roof inclination; The height of the building, the width length etc. and the many other such things where it becomes necessary to use trigonometry. Architects... read more

In construction we need trignometry to calculate the following:

  • Measuring fields, lots and areas;
  • Making walls parallel and perpendicular;
  • Installing ceramic tiles;
  • Roof inclination;
  • The height of the building, the width length etc. and the many other such things where it becomes necessary to use trigonometry.
  • Architects use trigonometry to calculate structural load, roof slopes, ground surfaces and many other aspects, including sun shading and light angles

In physics, trigonometry is used to find the components of vectors, model the mechanics of waves (both physical and electromagnetic) and oscillations, sum the strength of fields, and use dot and cross products. Even in projectile motion you have a lot of application of trigonometry.

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Lesson Posted on 26 Mar CBSE/Class 10/Mathematics

CHAPTER-10 CIRCLES NCERT SOLUTIONS

Rahul S

I am an experienced, qualified teacher and tutor with over 4 years of experience in teaching Maths ,...

Exercise: 10.1 How many tangents can a circle have? Answer A circle can have infinite tangents. Fill in the blanks : (i) A tangent to a circle intersects it in ............... point(s). (ii) A line intersecting a circle in two points is called a ............. (iii) A circle... read more

 

Exercise: 10.1

 

  1. How many tangents can a circle have?

 

Answer

 

A circle can have infinite tangents.

 

  1.  Fill in the blanks :

(i) A tangent to a circle intersects it in ............... point(s).

(ii) A line intersecting a circle in two points is called a .............

(iii) A circle can have ............... parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called ............

 

Answer

 

(i) one

(ii) secant

(iii) two

(iv) point of contact

 

  1. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
    a point Q so that OQ = 12 cm. Length PQ is :

    (A) 12 cm            (B) 13 cm          (C) 8.5 cm         (D) √119 cm

 

Answer

 

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ

By Pythagoras theorem in ΔOPQ,

OQ2 = OP2 + PQ2
⇒ (12)= 52 + PQ2

⇒PQ2 = 144 - 25

⇒PQ2 = 119

⇒PQ = √119 cm

(D) is the correct option.

 

  1. Draw a circle and two lines parallel to a given line such that one is a tangent and the

other, a secant to the circle.

 

Answer

 

AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Exercise: 10.2

 

In Q.1 to 3, choose the correct option and give justification.

 

  1.  From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

   (A)  7 cm             (B) 12 cm            (C) 15 cm                (D) 24.5 cm

 

Answer

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)

By Pythagoras theorem in ΔOPQ,

OQ2 = OP2 + PQ2
⇒ (25)= OP2 + (24)2

⇒OP2 = 625 - 576

⇒OP2 = 49

⇒OP = 7 cm

The radius of the circle is option (A) 7 cm.


2.  In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
     (A) 60°              (B) 70°                (C) 80°             (D) 90°

 

Answer

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT  = 360°
⇒ ∠PTQ + 90° + 110° + 90°  = 360°

⇒ ∠PTQ = 70°

∠PTQ is equal to option (B) 70°.

 

  1.  If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to
     (A) 50°                (B) 60°               (C) 70°                  (D) 80°

 

Answer

 

OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°

In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB  = 360°
⇒ ∠AOB + 90° + 90° + 80°  = 360°

⇒ ∠AOB = 100°

Now,

In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)

Thus ∠POB = ∠POA

∠AOB = ∠POB + ∠POA

⇒ 2 ∠POA = ∠AOB

⇒ ∠POA = 100°/2 = 50°

∠POA is equal to option  (A) 50°

 

 

 

  1. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

 

Answer

 

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B

respectively.
Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,

∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º

From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.

Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.

 

 

 

  1.  Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

 

Answer

 

Let AB be the tangent to the circle at point P with centre O.

We have to prove that PQ passes through the point O.

Suppose that PQ doesn't passes through point O. Join OP.

Through O, draw a straight line CD parallel to the tangent AB.

PQ intersect CD at R and also intersect AB at P.

AS, CD // AB PQ is the line of intersection,

∠ORP = ∠RPA (Alternate interior angles)

but also,

∠RPA = 90° (PQ ⊥ AB) 

⇒ ∠ORP  = 90°

∠ROP + ∠OPA = 180° (Co-interior angles)

⇒∠ROP + 90° = 180°

⇒∠ROP = 90°

Thus, the ΔORP has 2 right angles i.e. ∠ORP  and ∠ROP which is not possible.

Hence, our supposition is wrong. 

∴ PQ passes through the point O.

 

  1.  The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

 

Answer

 

AB is a tangent drawn on this circle from point A.

∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA2 = AB+ BO2
⇒ 5= 4+ BO2
⇒ BO2 = 25 - 16
⇒ BO2 = 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.

Answer

Let the two concentric circles with centre O.

AB be the chord of the larger circle which touches the smaller circle at point P. 

∴ AB is tangent to the smaller circle to the point P.

⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA2 =  AP2 + OP2
⇒ 52 = AP2 + 32
⇒ AP2 = 25 - 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle                      bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
 ∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer


From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)

BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC

 

 

  1. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

 

Answer

 

We joined O and C

A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,

 ΔOQB  ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°

 

  1.  Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

 

Answer

Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º

∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

 

  1. Prove that the parallelogram circumscribing a circle is a rhombus.

 

Answer

 

ABCD is a parallelogram,
∴ AB = CD ... (i)
∴ BC = AD ... (ii)

From the figure, we observe that,

DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) =

     (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC ... (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC ... (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.

 

  1. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and

DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

 

Answer

 

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x

We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x

Now semi perimeter of circle s,
⇒ 2s = AB + BC + CA
         = x + 8 + 14 + 6 + x
         = 28 + 2x
⇒s = 14 + x

Area of ΔABC = √s (s - a)(s - b)(s - c)

                         = √(14 + x) (14 + x - 14)(14 + - x - 6)(14 + - x - 8)

                         = √(14 + x) (x)(8)(6)

                         = √(14 + x) 48 x ... (i)

also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

                                 = 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

                                 = 2×1/2 (4x + 24 + 32) = 56 + 4... (ii)

Equating equation (i) and (ii) we get,

√(14 + x) 48 = 56 + 4x

Squaring both sides,

48x (14 + x) = (56 + 4x)2

⇒ 48x = [4(14 + x)]2/(14 + x)

⇒ 48x = 16 (14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

x = 7 cm

Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm

 

  1.  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. 

 

Answer

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS

⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7

Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

 

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Lesson Posted on 08 Mar CBSE/Class 10/Mathematics CBSE/Class 12/Science/Mathematics

Honesty and Dedication helps in education and success in life.

Jagdish S.

Mathematics teaching for CBSE ICSE,PUC state CET,JEE IIT and Engineering maths ,CAT,MAT,International...

Honesty and Dedication helps in education and success in life. As per my 22 years of experience of teaching mathematics i would like to say to all student that we should be Honesty and Dedicated in study then we will be successful in life. If we earn money or marks by wrong way then we will not feel... read more

Honesty and Dedication helps in education and success in life.

As per my 22 years of experience of teaching mathematics i would like to say to all student that we should be Honesty and Dedicated in study then we will be successful in life. If we earn money or marks by wrong way then we will not feel happiness. Get less mark but with honesty that will make you feel happiness.

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Answered on 02 Apr CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Sushil K.

Work hard till your signature becomes an autograph

Let the marks of mathematics be x Than marks of science be 28-x now according to question (x+3), (28-x-4) The product of new marks will be 180 So, (x+3) (28-x-4)=180 24x-x^2+72-3x=180 x^2-21x+108=0 x^2-12x-9x+108=0 x(x-12)-9(x-12)=0 (x-9)(x-12)=0 (x-9)=0 than x=9 (x-12)=0than x=12 If 9 marks got... read more

Let the marks of mathematics be x 

Than marks of science be 28-x 

now according to question 

(x+3), 

(28-x-4)

The product of new marks will be 180

So, 

(x+3) (28-x-4)=180

24x-x^2+72-3x=180

x^2-21x+108=0

x^2-12x-9x+108=0

x(x-12)-9(x-12)=0

(x-9)(x-12)=0

(x-9)=0 than x=9

(x-12)=0than x=12

If 9 marks got in math than 19 marks got in science

If 12 marks got in math than 16 marks got in science 

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Answered on 25 Feb CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Salman Ahmad Siddiqui

Maths Tutor

=1/9 use formula cosec² x -cot² x =1
Answers 19 Comments
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Answered on 08 Mar CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

If tan|+sin|=m and tan|- sin|=n, show that (m2-n2)=4?mn

Aravind Kumar

Answers 2 Comments
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