CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Let the speed of stream be x Speed of boat in still water is 8km/hr Let upstream be 8 - x & Downstream be 8 + x Now, time taken by boat to go 22km downstream = 22 / (8 + x) hr Time taken by boat to go 15 km upstream = 15 / (8 + x) hr Total time taken by boat = 5 hr So, 22 / (8 + x) + 15 / (8 + x) = 5 22(8 + x) + 15 (8 + x) / (8 + x) (8 - x) = 5 176 - 22x + 120 + 15x = 5(64 - x^2) 296 - 7x = 320 - 5x^2 5x^2 - 7x - 24 5x^2 - 15x + 18x - 24 = 0 (5x + 8) (x - 3) = 0 x is not equal to -8/5 , x = 3 Therefore, speed of stream is 3km/hr

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Tanu Priya 7 years, 9 months ago

See the examples of NCERT

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Similar angle

Posted by Shahwaiz Alam 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Harsh Rajput 7 years, 9 months ago

I think it is similar triangle

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Chapter 13 surface area and volume all formulae

Posted by Ankit Parmar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Mayank Bhunia 7 years, 9 months ago

Check last page of chapter

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Hi guys are all prepated for exams

Posted by Devild Back 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Sakshi Thakran 7 years, 9 months ago

Yes

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Amanpreet Singh 7 years, 9 months ago

i m not

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Kiska kiska maths prepare ho gya

Posted by Ankit Gupta 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Devild Back 7 years, 9 months ago

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Anurag Mishra 7 years, 9 months ago

Mera

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Monalika Sarangi 7 years, 9 months ago

Mera bilkul nhi hua hai .....

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If sec¶=x+1/4x then prove that sec¶+tan¶=2x or 1/2x

Posted by Anish Jain 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Ra Cartin 7 years, 9 months ago

Tan2= sec2-1 =(x+1/4x)square-1 =x2+1/16x2-1/2 (by a2+b2=(a+b)2-2ab) We get, (x-1/4x)=tan Sec+tan=x+1/4x+x-14x =2x

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A tree of 6m height cast a 4m long shadow,At same time a flag post casts a shadow 50m long .the height of the post is

Posted by Surajkant Kumar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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Which book i need to prefer for board exam rd sharma or ncert... Mths

Posted by Shraddha Solanki 7 years, 9 months ago

CBSE > Class 10 > Mathematics

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The angle of elevation of a cloud from a point 200m above a lake is 30 and the angle of depression of the reflection of the cloud in lake is 60. Find the height of the cloud.

Posted by Pranjal Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

In {tex}\triangle{/tex}ADC,
{tex}\tan 30 ^ { \circ } = \frac { H - 200 } { x }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { H - 200 } { x }{/tex}
{tex}\Rightarrow \quad x = \sqrt { 3 } {/tex}(H - 200) m. ............(1)
In {tex}\triangle{/tex}ADF,
{tex}\tan 60 ^ { \circ } = \frac { H + 200 } { x }{/tex}
{tex}\sqrt { 3 } = \frac { H + 200 } { x }{/tex}
{tex}\sqrt3x=H+200{/tex}...............(2)
here H is height of cloud above the lake.

Substituting the value of x from (1) into (2) we get
3(H - 200) = H + 200
3H - 600 = H + 200
2H = 800
H = 400 m
So height of the cloud above the lake is 400 m.

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Express the hcf of 48 and 18 as linear combination

Posted by Devashish K 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let us find HCF of 48 and 18
{tex}48 = 18 \times 2 + 12{/tex}
{tex}18 = 12 \times 1 + 6{/tex}
{tex}12 = 6 \times 2 + 0{/tex}
Hence HCF (48, 18) = 6
Now, 6 = 18 - 12 {tex}\times{/tex} 1
6 = 18 - (48 - 18 {tex}\times{/tex} 2)
6 = 18 - 48 {tex}\times{/tex} 1 + 18{tex}\times{/tex}2
6 = 18 {tex}\times{/tex} 3 - 48 {tex}\times{/tex} 1
6 = 18 {tex}\times{/tex} 3 + 48 {tex}\times{/tex}(-1)
i.e., 6 = 18x + 48y ........ (1)
where x= 3, y = -1
{tex}\therefore{/tex} 6 = 18 {tex}\times{/tex} 3 + 48 {tex}\times{/tex} (-1)
= 18 {tex}\times{/tex} 3 + 48 {tex}\times{/tex} (-1) + 18 {tex}\times{/tex} 48 - 18 {tex}\times{/tex} 48
= 18(3 + 48) + 48(-1 - 18)
= 18 {tex}\times{/tex} 51 + 48 {tex}\times{/tex} (-19)
6 = 18x + 48y ...... (2)
where x = 51, y = -19
Hence, x and y are not unique.

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What is the probability that a non leap year has 53 monday

Posted by Abhishek Verma 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Jyoti Vaibhaw 7 years, 9 months ago

1/7

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Varsha Saini 7 years, 9 months ago

Non leap year has 365 days that is 364 days +1 day extra so the extra day can be Monday, Tuesday, Wednesday... Sunday total outcome are 7 , favorable outcome that is Monday is 1 so probability that a non leap year has 53 Mondays is 1/7.

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Sourav Gujjar 7 years, 9 months ago

53/365

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Deepanshu Maheshwari 7 years, 9 months ago

2 /7

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7/50 find decimal expansion without long division

Posted by Aakash Rajoriya 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Cutiee Princess 7 years, 9 months ago

7 / 2*5*5 = 7*2/2*2*5*5= 14/100=0.14

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Hlo everyone ? ? gud mrng? ?

Posted by Eva Maijoli 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Anurag Mishra 7 years, 9 months ago

Fine

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Jyotsana Singh 7 years, 9 months ago

Good morning how are you

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Anurag Mishra 7 years, 9 months ago

Hlo, gm

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Sourav Gujjar 7 years, 9 months ago

Gud morning

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If Tan A = n tanB and sinA=m sinB prove that cosA =m-1/m+1

Posted by Shahana Aziyaa Azmi 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given,
tan A = n tan B

{tex} \Rightarrow{/tex} tanB = {tex}\frac{1}{n}{/tex}tan A

{tex}\Rightarrow{/tex} cotB = {tex}\frac { n } { \tan A }{/tex}..........(1)

Also given,

sin A = m sin B

{tex}\Rightarrow{/tex} sin B = {tex}\frac{1}{m}{/tex}sin A

{tex}\Rightarrow{/tex} cosec B = {tex}\frac { m } { \sin A }{/tex}.....(2)

We know that, cosec²B - cot²B = 1, hence from (1) & (2) :-

{tex} \quad \frac { m ^ { 2 } } { \sin ^ { 2 } A } - \frac { n ^ { 2 } } { \tan ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow \quad \frac { m ^ { 2 } } { \sin ^ { 2 } A } - \frac { n ^ { 2 } \cos ^ { 2 } A } { \sin ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow \quad \frac { m ^ { 2 } - n ^ { 2 } \cos ^ { 2 } A } { \sin ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow{/tex} m² - n²cos²A = sin²A
{tex}\Rightarrow{/tex} m² - n²cos²A = 1 - cos²A
{tex}\Rightarrow{/tex} m² - 1 = n²cos²A - cos²A
{tex}\Rightarrow{/tex} m² - 1 = (n² - 1) cos²A
{tex}\Rightarrow \quad \frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex} cos²A