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Trigonometry is a branch of mathematics that studies relationships involving lengths and angles of triangles. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. This can only be understood by learning and practing......???

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Volume of cone height 13 radius 4

Posted by Rashid ¥€$ 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Satyam Meher 7 years, 1 month ago

It is wrong

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Nikhil Singh Pratap 7 years, 1 month ago

By the help of1/3×22/7 ×16 ×13=217.90

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A coin is tossed once.what is the probability of getting tail?

Posted by Piyush Chaurasia 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Aditya Sahu 7 years, 1 month ago

1/2

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Shalu Gupta 7 years, 1 month ago

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NCERT examplar math guide important question of chapter 6th.

Posted by Rihina Sumbhania 7 years, 1 month ago

CBSE > Class 10 > Mathematics

0 answers

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For what value of x the points A(1,2),B(-4,7)andC(x,-1) are collinear?

Posted by Swastika Sharma 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Ankit Jaiswal 7 years, 1 month ago

X,0

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If areas of two similar triangles are equal.prove that they are congruent.

Posted by Praveen Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given : {tex}\triangle \mathrm{ABC}{\sim} \triangle \mathrm{PQR}{/tex} &
ar {tex}\triangle \mathrm{ABC}{/tex} = ar {tex}\Delta \mathrm{PQR}{/tex}
{tex}{/tex}To prove: {tex}\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}{/tex}
Since, {tex}\triangle \mathrm{ABC}{\sim} \Delta \mathrm{PQR}{/tex}
ar {tex}\triangle A B C=\text { ar } \triangle P Q R{/tex} (given)
{tex}\frac{{\Delta ABC}}{{ar\Delta PQR}} = 1{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{{A{B^2}}}{{P{Q^2}}} = \frac{{B{C^2}}}{{Q{R^2}}} = \frac{{C{A^2}}}{{P{R^2}}} = 1{/tex}
[Using Theorem of area of similar Triangles]
{tex}\Rightarrow{/tex} AB = PQ, BC = QR & CA = PR
Thus, {tex}\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}{/tex}

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Find the roots. (x-5)(x-6)=25/24²

Posted by Saurav Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

According to the question,
{tex}(x - 5)(x - 6) = \frac{{25}}{{{{\left( {24} \right)}^2}}}{/tex}
{tex}\Rightarrow x(x - 6) - 5(x - 6) = \frac{{25}}{{{{(24)}^2}}}{/tex}
{tex}\Rightarrow {x^2} - 6x - 5x + 30 - \frac{{25}}{{{{(24)}^2}}} = 0{/tex}
{tex}\Rightarrow {x^2} - 11x + 30 - \frac{{25}}{{{{(24)}^2}}} = 0{/tex}
{tex}\Rightarrow {x^2} - 11x + \frac{{30 \times {{24}^2} - 25}}{{{{(24)}^2}}} = 0{/tex}
{tex} \Rightarrow {x^2} - 11x + \frac{{30 \times 576 - 25}}{{{{(24)}^2}}} = 0{/tex}
{tex} \Rightarrow {x^2} - 11x + \frac{{17280 - 25}}{{{{(24)}^2}}} = 0{/tex}
{tex}\Rightarrow {x^2} - \frac{{264x}}{{24}} + \frac{{145}}{{24}} \times \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow {x^2} - \left( {\frac{{145}}{{24}} + \frac{{119}}{{24}}} \right)x + \frac{{145}}{{24}} \times \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow {x^2} - \frac{{145}}{{24}}x - \frac{{119}}{{24}}x + \frac{{145}}{{24}} \times \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow x\left( {x - \frac{{145}}{{24}}} \right) - \frac{{119}}{{24}}\left( {x - \frac{{145}}{{24}}} \right) = 0{/tex}
{tex}\Rightarrow \left( {x - \frac{{145}}{{24}}} \right)\left( {x - \frac{{119}}{{24}}} \right) = 0{/tex}
{tex}\Rightarrow x - \frac{{145}}{{24}} = 0{/tex} or {tex}x - \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow x = \frac{{145}}{{24}}{/tex} or {tex}x = \frac{{119}}{{24}}{/tex}

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(M/3+3/n)^2-2mn

Posted by Kshitij Tyagi 7 years, 1 month ago

CBSE > Class 10 > Mathematics

0 answers

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What is refractive index

Posted by Kkk Kb 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

????? ??? 7 years, 1 month ago

In this answer sin i upon m hai sin r

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????? ??? 7 years, 1 month ago

The value of the constant sin I upon me sin r for a ray of light passing from air into a particular medium is called the refractive index

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OACB is a quafrant of a circle with centre O and radius 3.5cm. If OD=2 cm, find the area of the (1) quadrant OACB, (2) shaded region.

Posted by Biwa Bisht 7 years, 1 month ago

CBSE > Class 10 > Mathematics

0 answers

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2+3÷5÷43÷2

Posted by Biwa Bisht 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Abhishek Bhardwaj 7 years, 1 month ago

2.0069

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Construct an isosceles triangle whose base is 8 cm and altitude 5cm and then construct

Posted by Arjun Haridas 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Arghyadeep Kolay 7 years, 1 month ago

First draw the base AB=8cm..... Then on the perpendicular bisector of AB, Cut CD=5cm i.e. D is the point at which AB is bisected and D is a point on the perpendicular bisector ...... Join AC and BC

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The height of Tower is 10 metre what is the length of its Shadow when sun's altitude is 45degree?

Posted by Prajwal Menasagi 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Tushar Kumar 7 years, 1 month ago

10 met

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3x^2 -- 2root6x + 2 = 0

Posted by Anjali Kumari??? 7 years, 1 month ago

CBSE > Class 10 > Mathematics

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A rational number can be expressed as a terminating decimal if the denominator has the factor (A)2or 5 (B)2and 5

Posted by Ujjawal Singh Rajput 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Janvi Bana 7 years, 1 month ago

option A

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Navya Singh 7 years, 1 month ago

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(tanA + cotA)2 - (tanA - cotA)2 = 4

Posted by Ragi Cherry 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Kunal J 7 years, 1 month ago

(tanA + cotA)^2 - (tanA - cotA)^2 tan^2 + cotA^2 + 2tanAcotA - (tanA^2 +cotA^2 - 2tanAcotA) tan^2 + cotA^2 + 2 - tanA^2 - cotA^2 +2 = 2+2 = 4 Note tanA × cotA =1 Since tanA = 1/cotA or vice versa

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The question of chapter 9

Posted by M I 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Check NCERT solutions here : <a href="https://mycbseguide.com/ncert-solutions.html">https://mycbseguide.com/ncert-solutions.html</a>

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What substation mathod

Posted by Aditya Sah 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Guru Jatt 7 years, 1 month ago

Substitution is a method in which first put out the temporary value of variable then put in equation after this the exact value of variable is comes out then put in another equation.

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