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Jatin Kushwaha 15 hours ago

If two distances are equal and the points are not collinear then points form a isosceles triangle.
To prove that it is also right triangle , use the converse of Pythagoras
Check that is square of longest sides equal to sum of square of two shorter side.
H^2=P^2+B^2

Posted by Taheer Khan Taheer Khan 15 hours ago

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Jatin Kushwaha 14 hours ago

Napoleon code was implemented by Napoleon .
It abolished the fedual system privileges based on birth and established equality before law and provide right to property.

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Sumit Yadav 16 hours ago

A(-1,0)B(3,1)C(2,2)D(x, y) ABCD is a diagonal apply mid point method AC and BD and 4th vertex (-2,1)

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Aditya Choudhary 19 hours ago

Yes, there are some parents like Mrs Pumphrey who spoil their children by over-indulging and pampering them. They try to overfeed their children or pets despite knowing the ill-effects of overeating on their health.
Mark me perfect ...

Manveet Singh Gurjar 20 hours ago

Ya there are parents like Mrs. Pumphery.
But she is a pet owner in chapter they use to talk about pet his name was Tricki.

Posted by Vanshika Kaur 20 hours ago

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Sumit Yadav 16 hours ago

the y axis divides the line joining the points (-2, -3) and (3, 7) be k : 1. point of intersection line to y axis to be (0, y). Apply section formula and answer 2:3

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The Mystery 20 hours ago

Learn more:

[tex]\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}[/tex]

[tex]\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}[/tex]

The Mystery 20 hours ago

$$sinA = x$$

$$secA=y$$

$$\sf \bf :\longmapsto \dfrac{1}{cosA} = y$$

$$\sf \bf :\longmapsto \dfrac{sinA}{cosA} = xy$$

$$\sf \bf :\longmapsto tanA =xy$$

Learn more:

$$\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf {\angle A} & \{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &{ \sf{90}^{ \circ}} \\ \hline \\ \rm {sin A} & \green{0} & {\dfrac{1}{2}}& {\dfrac{1}{ \sqrt{2} }} &{ \dfrac{ \sqrt{3}}{2} }&{1} \\ \hline \\ \rm {cos \: A} & {1} &{ \dfrac{ \sqrt{3} }{2}}&{ \dfrac{1}{ \sqrt{2} }} & {\dfrac{1}{2}} &{0} \\ \hline \\\rm {tan A}& {0} &{ \dfrac{1}{ \sqrt{3} }}&{1} & {\sqrt{3}} & \rm {\infty} \\ \hline \\ \rm {cosec A }& \rm {\infty} & {2}& {\sqrt{2} }&{ \dfrac{2}{ \sqrt{3} }}&{1} \\ \hline\\ \rm {sec A} & {1 }&{ \dfrac{2}{ \sqrt{3} }}& {\sqrt{2}} & {2} & \rm {\infty} \\ \hline \\ \rm {cot A }& \rm {\infty} & {\sqrt{3}}& {1} & {\dfrac{1}{ \sqrt{3} }} &{0}\end{array}}}}$$

$$secA=y$$

$$\sf \bf :\longmapsto \dfrac{1}{cosA} = y$$

$$\sf \bf :\longmapsto \dfrac{sinA}{cosA} = xy$$

$$\sf \bf :\longmapsto tanA =xy$$

Learn more:

$$\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf {\angle A} & \{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &{ \sf{90}^{ \circ}} \\ \hline \\ \rm {sin A} & \green{0} & {\dfrac{1}{2}}& {\dfrac{1}{ \sqrt{2} }} &{ \dfrac{ \sqrt{3}}{2} }&{1} \\ \hline \\ \rm {cos \: A} & {1} &{ \dfrac{ \sqrt{3} }{2}}&{ \dfrac{1}{ \sqrt{2} }} & {\dfrac{1}{2}} &{0} \\ \hline \\\rm {tan A}& {0} &{ \dfrac{1}{ \sqrt{3} }}&{1} & {\sqrt{3}} & \rm {\infty} \\ \hline \\ \rm {cosec A }& \rm {\infty} & {2}& {\sqrt{2} }&{ \dfrac{2}{ \sqrt{3} }}&{1} \\ \hline\\ \rm {sec A} & {1 }&{ \dfrac{2}{ \sqrt{3} }}& {\sqrt{2}} & {2} & \rm {\infty} \\ \hline \\ \rm {cot A }& \rm {\infty} & {\sqrt{3}}& {1} & {\dfrac{1}{ \sqrt{3} }} &{0}\end{array}}}}$$

Prajan Elango 21 hours ago

Tan A = SinA/Cos A
Sec A = 1/Cos A
So
TanA = SinA * Sec A = x*y = xy

Posted by Alok Gameing 22 hours ago

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Posted by Rudra Prasad 1 day, 1 hour ago

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Ishitha Ishitha 42 minutes ago

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