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Secant 30 degree + cosec 30 degrees

Posted by Fathima Fathima 4 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Tanisha Das 4 years, 8 months ago

sec30°+cosec30° = 2/√3 + 2 =2+2√3/√3 =2+2√3/√3 x √3/ √3 =2√3+6/3 =2√3+2

Yash Bansal 4 years, 8 months ago

Sec30°= 1/2 Cosec30°= 2 So, Sec30° + Cosec30°=1/2 + 2 =1/2 + 2/1 =1 + 4/2 =5/2

ANSWER

Seema placed a light bulb at a point O on the ceiling and deirectly below it placed a table. Now she put a cardboard of shape ABCD between table and lighted bulb. Then a shadow of ABCD is cast on the table as A|B |C |D | . Quadrilateral A |B |C |D | is an enlargement of ABCD with scale factor 1 : 2. Also, AB= 1.5 cm, BC= 2.5 cm, CD= 2.4 cm and AD= 2.1 cm, A=105 , B= 100 , C= 70 and D= 85

Posted by Chaithanya Paul 4 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

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

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Yash Bansal 4 years, 8 months ago

1) Draw a perpendicular bisector of AB. Mark the point as X. 2) Taking XB as a radius, draw a dotted circle from X. 3) Join the points with B where the dotted circle cuts the circle (with A as centre). Do the Same for other circle, taking XA as radius

ANSWER

Find a Quadratic polynomial whose zeroes are 2- ? and 2+ ?

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Yash Bansal 4 years, 8 months ago

Sum of zeroes= (2-3)+(2+3)= 2+2=4........ Product of zeros=(2-3)(2+3)= 4+6-6-9 = -5 ....... Using formula, x^2+(Sum of zeroes× x) + (product of zeroes) = x^2+4x-5

Bharti Bharadwaj 4 years, 8 months ago

x^2 - 4x +1

ANSWER

Find a Quadratic polynomial whose zeroes are 2- ? and 2+ ? 2

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Aryan Tiwari 4 years, 8 months ago

answer is x2-33x-22

ANSWER

Find the point on the x-axis which is equidistant from (2,-5) and (-2, 9)

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

Now, the point is on x-axis, so the value 'y' of the given point will be zero.

Let, the value of the 'x' value of the given point = x

[Assume,x as a variable to do the further mathematical calculations.]

So,the point is = (x,0)

As mentioned in the question,the two points are equidistant from (x,0).

So,

Distance between (x,0) and (2,-5)

= ✓(x-2)²+(0+5)²

Distance between (x,0) and (-2,9).

=✓(x+2)²+(0-9)²

Now,the points are equidistant.

So,

✓(x-2)²+(0+5)² = ✓(x+2)²+(0-9)²

(x-2)²+(0+5)² = (x+2)²+(0-9)²

x²-4x+4+25 = x²+4x+4+81

x²-4x-x²-4x = 4+81-4-25

-8x = 56

x = -7

ANSWER

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds. If they change simultaneously at 7 AM, after what time will they change again simultaneously

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

The traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds, respectively.

So let us take the LCM of the given time that is 48 seconds, 72 seconds, 108 seconds

⇒ 48 = 2 × 2 × 2 × 2 × 3

⇒ 72 = 2 × 2 × 2 × 3 × 3

⇒ 108 = 2 × 2 × 3 × 3 × 3

Hence, LCM of 48, 72 and 108 = (2 × 2 × 2 × 2 × 3 × 3 × 3)

LCM of 48, 72 and 108 = 432

So after 432 seconds, they will change simultaneously

We know that

60 seconds = 1 minute

so on dividing 432 / 60, we get 7 as quotient and 12 as a reminder

Hence, 432 seconds = 7 min 12 seconds

∴ The time = 7 a.m. + 7 minutes 12 seconds

Hence, the lights change simultaneously at 7:07:12 a.m

ANSWER

In the figure, Quadrilateral ABCD is circumscribing a circle with centre O and AD⊥ AB. If the radius of incircle is 10 cm, then the value of x is

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

Let O be the mid point,

As = Ap ( tangent to the given circle)

and since L SAp is 90°

As = OP = Ap = 10cm ( radius)

Now, Let t be the point where Ap meets Cq

and we know that Cq = PR = 27 (tangent)

and Ct = 38 ( given in the fig)

» Cq + qt = 38

» 27cm + qt = 38cm

» qt = 38cm-27cm = 11cm

and now tp = tq ( tangent)

tp = 11cm

So to find x , i.e Apt

we have Apt = Ap + pt

= 10cm + 11cm = 21 cm

ANSWER

Prove that 5 - ? is irrational given that ? is irrational

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Bhoomika Verma 4 years, 8 months ago

Let as assume that 5-√3 is rational no. 5-√3 =a/b ( a and b is coprime no. & Integers &q≠0) 5-a/b = √3 √3= 5b-a/ b Since a and b are integers ,we get 5-a/ b is rational no. But √3 is irrational no. So, our assumption was wrong . Hence 5-√3 is irrational no.

Yash Bansal 4 years, 8 months ago

Lets us assume that 5 is rational! So, subtraction of rational with irration results as irrational, so, 5-3 is irrational

ANSWER

one root of the quadratic equation 2x2 –px+3=0 is ? ? then find the value of p and the other root of the equation

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

ANSWER

n an equilateral ABC, D is a point on side BC such that BD =? ? BC prove that 9AD2 = 7AB

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .

Now, ∆ABE and ∆AEC
∠AEB = ∠ACE = 90°
AE is common side of both triangles ,
AB = AC [ all sides of equilateral triangle are equal ]
From R - H - S congruence rule ,
∆ABE ≡ ∆ACE
∴ BE = EC = BC/2

Now, from Pythagoras theorem,
∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)
∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)

From equation (1) and (2)
AB² - AD² = BE² - DE²
= (BC/2)² - (BE - BD)²
= BC²/4 - {(BC/2) - (BC/3)}²
= BC²/4 - (BC/6)²
= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9
∵AB = BC = CA
So, AB² = AD² + 2AB²/9
9AB² - 2AB² = 9AD²
Hence, 9AD² = 7AB²

ANSWER

Prove the following Identity ???? ?+???? + ?+???? ???? = ????? 3

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

Given cosA/(1+sinA) +(1+sinA)/cosA

On taking the LCM we get,

={cos²A +(1+sinA)²}/cosA.(1+sinA)

Combining the like terms and adding and also we know that,use sin²A+cos²A =1

=(1+1+2sinA)/cosA(1+sinA)

=2(1+sinA)/cosA(1+sinA)

=2/cosA

=2secA

ANSWER

Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of m for which y = mx + 3

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Nargish Akhtar 4 years, 8 months ago

Let, P(x)=2x+3y=11-(1) P(x)=2x-4y=-24-(2) Now solving 1 and 2 we get 7y=35 Y=5 Now putting the value of y in 1 we get X=-2 Now,given y=MX+3 5=m(-2)+3 M=-1

Himansh Makkar 4 years, 8 months ago

2x + 3y = 11 (1) 2x - 4y = -24 (2) Subtracting both the equations 7y = 35 y = 5 Putting y = 5 in 1 we get 2x + 15 = 11 2x = -4 x = -2 y = mx + 3 5 = m(-2) + 3 2 = -2m -1 = m

ANSWER

The angle of elevation of the top of a tower from a point on the ground which is 30m away from the foot of the tower is 300 . Find the height of the tower

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Kiran Kumari 4 years, 8 months ago

Recheck your question

ANSWER

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time; a tower casts a shadow 28m. Find the height of the tower

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Saniya Khanam 4 years, 8 months ago

Height of pole=AB=6 m Length of shadow of pole =BC=4 m Length of shadow of tower=EF=28 m In △ABC and △DEF ∠B=∠E=90∘ both 90∘ as both are vertical to ground ∠C=∠F (same elevation in both the cases as both shadows are cast at the same time) ∴△ABC∼△DEF by AA similarity criterion We know that if two triangles are similar, ratio of their sides are in proportion So,DEAB=EFBC ⇒DE6=284 ⇒DE=6×7=42 m Hence the height of the tower is 42 m I hope it will help you??

ANSWER

Two poles of equal heights are standing opposite each other on either side of the road which is 80m wide. From a point O between them on the road, the angles of elevation of the top of the poles are 600 and 300 respectively. Find the height of the poles and the distances of the point from the poles.

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

ANSWER

Show that the points (1,7), (4,2), (-1,-1) and (-4,4) are the vertices of a square

Posted by C Akhilesh 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

ANSWER

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water

Posted by C Akhilesh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

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

Posted by Pihu ✨ 4 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 3 months ago

Let ABCD be a quadrilateral circumscribing a circle with centre O.
Now join AO, BO, CO, DO.
From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]
Let ∠DAO=∠BAO=1
Also ∠ABO=∠CBO [Since, BA and BC are tangents]
Let ∠ABO=∠CBO=2
Similarly we take the same way for vertices C and D
Sum of the angles at the centre is 360^o
Recall that sum of the angles in quadrilateral, ABCD = 360^o

= 2(1+2+3+4)=360^o
=1+2+3+4=180^o
In ΔAOB, ∠BOA=180−(1+2)
In ΔCOD, ∠COD=180−(3+4)
∠BOA+∠COD=360−(1+2+3+4)

=360^o–180^o
=180^o

Since AB and CD subtend supplementary angles at O.
Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

ANSWER

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

Posted by Pihu ✨ 4 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Kiran Chaudhri 4 years, 8 months ago

Let AT be the tangent drawn from a point A to a circle with centre O and OA=5 cm and AT=4 cm. Since tangent at a point is perpendicular to the radius through the point of contact ∴OT⊥AT ∴ from right angled △OAT, (OA)2=(OT)2+(TA)2 ⇒(5)2=(OT)2+(4)2 ⇒25−16=(OT)2 ⇒9=(OT)2 ⇒OT=3 cm ∴ radius of the circle =3 cm.

ANSWER

Cards mark between the number 2 to 101 are placed in a box and mixed thoroughly one card is drawn from this box find the probability that the number on the card is first on number less than 10 a number which is a perfect square

Posted by Punit Patil 4 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

1/2a+b+2x = 1/2a +1/b +1/2x find the roots

Posted by Charvi Sangwan 4 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Ex 8.3 cut ho gyi h kya

Posted by Misty Chan 4 years, 8 months ago

CBSE > Class 10 > Mathematics

5 answers

Rahul Rajput 4 years, 8 months ago

Haa

Misty Chan 4 years, 8 months ago

Therefore, my prediction was right Thank you all for answering!!!

Prem Kumar 4 years, 8 months ago

Yes, but only in 2021 in our exam.

Siya Singhal 4 years, 8 months ago

Ha

Rachit Dwivedi 4 years, 8 months ago

Haa

ANSWER

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height, and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)

Posted by Neelam Soring 4 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Sadiya Ali 4 years, 8 months ago

Area to be colored = CSA OF HEMISPHERE +CSA OF CONE Radius of hemisphere and cone =7\4 Height of cone =15-7/4 Calculation is your part???

Mayank Dagar 4 years, 8 months ago

Formula = 2*22/7*r*r

ANSWER

2×0

Posted by Kunal Singh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

5 answers

Yd Sharma 4 years, 8 months ago

0

Krishna Kumar 4 years, 8 months ago

0

Good Day 4 years, 8 months ago

0

46B Urmee Verma 4 years, 8 months ago

0

Bharani Daran.R 4 years, 8 months ago

0

ANSWER

What is 1-sin square Q

Posted by Hirender Singh 4 years, 8 months ago

CBSE > Class 10 > Mathematics

4 answers

Mayank Dagar 4 years, 8 months ago

Cos square Q

Siya Singhal 4 years, 8 months ago

Cos square Q

Good Day 4 years, 8 months ago

Yaha par Q kya hai

...... ...... 4 years, 8 months ago

Cos squareQ

ANSWER

Find the HCF & LCM of 404 & 96

Posted by Sarah Mariyam 4 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Good Day 4 years, 8 months ago

HCF = 4 LCM = 9696

Sumesh ☺️☺️☺️ 4 years, 8 months ago

HCF IS 4 AND LCM IS 9696

ANSWER

if the mode of the data : 3,5,8,9,12,7,12and x is 8,find the value of x

Posted by Aarthi Sivanthirajan 4 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Shreyansh Sharma 4 years, 8 months ago

8

ANSWER

If the pair of equations kx - 2y = 3 and 3x + y 5 represent two intersecting lines, then find the value of k

Posted by Jhon Rathi 4 years, 8 months ago

CBSE > Class 10 > Mathematics

3 answers

Bhoomika Verma 4 years, 8 months ago

K≠ -6

Rajni Garg 4 years, 8 months ago

K is not equal to-6

Sumesh ☺️☺️☺️ 4 years, 8 months ago

Can you check your question again

ANSWER

If the pair of equations kx - 2y = 3 and 3x + y 5 represent two intersecting lines, then find the value of k. =

Posted by Jhon Rathi 4 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

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