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

We have, AB = 16 cm
{tex} \therefore{/tex} AL = BL = 8 cm [Perpendicular from the centre of the circle to a chord bisects it]
In {tex}\triangle{/tex}OLB, we have,
OB² = OL² + LB²
{tex} \Rightarrow{/tex} 10² = OL² + 8²
{tex} \Rightarrow{/tex} OL² = 100 - 64 = 36
{tex} \Rightarrow{/tex} OL = 6 cm
Let PL = x and PB = y. Then, OP = (x + 6) cm.
Now, {tex} \angle PBO = 90^\circ{/tex} as tangent makes a right angle with the radius of the circle at the point of contact.
In {tex}\triangle{/tex}'s PLB and OBP, we have,
PB² = PL² + BL² and OP² = OB² + PB²
{tex} \Rightarrow{/tex} y² = x² + 64 and (x + 6)² = 100 + y² [Substituting the value of y² in second equation]
{tex} \Rightarrow{/tex} (x + 6)² = 100 + x² + 64
{tex} \Rightarrow{/tex} 12x = 128
{tex}\Rightarrow x = \frac{{32}}{3}cm{/tex}
{tex} \therefore{/tex} y² = x² + 64
{tex} \Rightarrow {y^2} = {\left( {\frac{{32}}{3}} \right)^2} + 64 = \frac{{1600}}{9}{/tex}
{tex} \Rightarrow y = \frac{{40}}{3}cm{/tex}
Therefore, {tex} PA = PB = \frac{{40}}{3}cm{/tex}

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A ladder rests against a wall at an angle alpha to horizontal. its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle betta with the horizontal. Show that A=cos alpha_ cos betta/b=sin betta _ son alpha

Posted by Neha Upadhyay 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let PQ be the ladder such that is top Q is on the wall OQ.
The ladder is pulled away from the wall through a distance a, so Q slides and takes position Q'.
Clearly, {tex}PQ = P'Q'.{/tex}
In {tex}\Delta 's{/tex} {tex}POQ \ and \ P'OQ', {/tex}we have
{tex}\sin \alpha = \frac{{OQ}}{{PQ}},\cos \alpha = \frac{{OP}}{{PQ}},\sin \beta = \frac{{OQ'}}{{P'Q'}},\cos \beta = \frac{{OP'}}{{P'Q'}}{/tex}
{tex} \Rightarrow \sin \alpha = \frac{{b + y}}{{PQ}},\cos \alpha \frac{x}{{PQ}},\sin \beta = \frac{y}{{PQ}},\cos \beta = \frac{{a + x}}{{PQ}}{/tex}
{tex} \Rightarrow \sin \alpha - \sin \beta = \frac{{b + y}}{{PQ}} - \frac{y}{{PQ}}{/tex} and
{tex}\cos \beta - \cos \alpha = \frac{{a + x}}{{PQ}} - \frac{x}{{PQ}}{/tex}
{tex} \Rightarrow \sin \alpha - \sin \beta = \frac{b}{{PQ}}{/tex} and
{tex}\cos \beta - \cos \alpha = \frac{a}{{PQ}}{/tex}
{tex} \Rightarrow \frac{a}{b} = \frac{{\cos \alpha - \cos \beta }}{{\sin \beta - \sin \alpha }}{/tex}

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tan A/SecA-1+ tanA/ sec A+1=2cosecA

Posted by Archana Sharma 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Aman Kumar 7 years, 9 months ago

Ho jayega it's easy

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Exercises 4.3 question no 8 how to solve in easy method or type

Posted by Meenakshi Bhardwaj 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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10 years ago the ages of two sons was one third of their fathers age. One son is two years older than other and sum of their present ages is 14 years less than fathers present age. Find the present ages of all.___#only expert answer.

Posted by Vipul Pandey 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Vipul Pandey 7 years, 9 months ago

Father age 46 One son age 15 another 17

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Sourav Soni 7 years, 9 months ago

2.5/4.5 (sons age) 21(father age)

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Sin a + cos a

Posted by S K 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Aman Kumar 7 years, 9 months ago

Hi sona

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Keshav Chauhan 7 years, 9 months ago

Incomplete question

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Aman Kumar 7 years, 9 months ago

Complete the question

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Arrre kis kis ko dar lag raha hai 10 class board se???

Posted by Rohan Singh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

7 answers

Rohan Singh 7 years, 9 months ago

Shi kha friends...

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Akshit Saini 7 years, 9 months ago

Bhai darna nhi h confidence se paper do

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Miss Universe 7 years, 9 months ago

Kaisa dar just chill guys

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Aman Pal 7 years, 9 months ago

Kyu*

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Aman Kumar 7 years, 9 months ago

Ab ye bol k kya fayada. Board toh Dena hi h

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Aman Pal 7 years, 9 months ago

Kyi darna ?????

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R_ Sandhu_ 7 years, 9 months ago

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Find the volume of a solid in the form of a right circular cylinder with hemisphere ends, whose total height is 2.7m and the diameter of each hemispherical end is 0.7m.

Posted by Avanish Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Diameter of common base = 0.7m
Radius of common base = {tex}\frac{0.7}{2}{/tex}m
Total height of solid = 2.7m
Height of cylinder = 2.7 - {tex}\frac{0.7}{2}{/tex} - {tex}\frac{0.7}{2}{/tex} = 2 m
{tex}\therefore{/tex} Volume of solid = volume of cylinder + 2 {tex}\times{/tex}volume of hemisphere
{tex}= \pi r ^ { 2 } h + 2 \times \frac { 2 } { 3 } \pi r ^ { 3 }{/tex}
{tex}= \frac { 22 } { 7 } \times \frac { 0.7 } { 2 } \times \frac { 0.7 } { 2 } \times 2 + 2 \times \frac { 2 } { 3 } \times \frac { 22 } { 7 } \times \frac { 0.7 } { 2 } \times \frac { 0.7 } { 2 } \times \frac { 0.7 } { 2 }{/tex}
{tex}= \frac { 22 } { 7 } \times \frac { 0.7 } { 2 } \times \frac { 0.7 } { 2 } \times 2 \left[ 1 + \frac { 2 } { 3 } \times \frac { 0.7 } { 2 } \right]{/tex}
{tex}= \frac { 22 } { 7 } \times \frac { 0.7 } { 2 } \times \frac { 0.7 } { 2 } \times 2 \times \frac { 3.7 } { 3 }{/tex}
=0.949 m³
=0.95 m³ (approx.)