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Jatin Kushwaha 17 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

No solution

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)

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

The Mystery 22 hours ago

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

The Mystery 22 hours ago

$$sinA = x$$
$$secA=y$$
$$\sf \bf :\longmapsto \dfrac{1}{cosA} = y$$
$$\sf \bf :\longmapsto \dfrac{sinA}{cosA} = xy$$
$$\sf \bf :\longmapsto tanA =xy$$

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

Prajan Elango 23 hours ago

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

🤟Royal Thakur🤟 1 day, 17 hours ago

😳😳Wow bro 🙌🙌🙌🙌🙌🙌

The Mystery 1 day, 17 hours ago

Edited table :

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

The Mystery 1 day, 17 hours ago

$$\sf \bf :longmapsto xcosA = 8$$
$$\sf \bf :longmapsto cosA=\dfrac{8}{x}$$
Taking second equation,
$$\sf \bf :longmapsto 15cosecA = 8secA$$
$$\sf \bf :longmapsto 15\dfrac{1}{sinA} = 8\dfrac{1}{cosA}$$
$$\sf \bf :longmapsto tanA = \dfrac{15}{8}$$

Thus, you can see using a triangle that sides are respectively, 15k, 8k, 17k
$$\sf \bf :longmapsto cosA = \dfrac{8}{17}$$
Hence , x = 17

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

The Mystery 1 day, 17 hours ago

17

The Mystery 1 day, 2 hours ago

How 6 aren't 5? And total are 50 not 54

Its Nav Sandhu ✌✌ 1 day, 6 hours ago

6 to 50 there were 6 perfect square and total no. 54 so P (E) :- 6/54 = 3/27 = 1/9 answer

The Mystery 1 day, 16 hours ago

Perfect square number from 1 to 50 are only 5 (as 7²=49)
$$\sf \bf :\longmapsto P = \dfrac{5}{50} = \dfrac{1}{10}$$

Kunal Chauhan 1 day, 23 hours ago

Hi

Bhavya Maheshwari 1 day, 21 hours ago

Tan theta = 3/4 then P=3k B=4k H=5k ( Pythagoras) Cos² - sin² = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25

Reena Ahirwar 2 days, 5 hours ago

Therefore we can say that when tan theta =34,cos 2 theta -sin 2 theta =725

Sakshi Kumari 2 days, 4 hours ago

45×2 = 90

X-O Harsh 2 days, 16 hours ago

(45)^2=90

Aastha Lilhore 2 days, 17 hours ago

45*2 = 90

Its Nav Sandhu ✌✌ 1 day, 20 hours ago

D. 2520 Its a comman and very important ques so never forget him

Prateek Aggarwal 2 days, 16 hours ago

D 2520

Monarch Pathak¹³ 2 days, 18 hours ago

By using Euclid's division lemma (a=bq+r) Where A=225. B=135 Step 1 225=135×1+90 Step 2 A=135. B=90 135=90×1+45 Step 3 A=90. B=45 90=45×2+0 So our HCF of (135 and 225) is 45 Hence proved

Subodh Kumar 3 days, 17 hours ago

Hi

Rohan Kandulna 3 days, 19 hours ago

Questions no 1

Rohan Kandulna 3 days, 19 hours ago

Experience 7.1

Abbhiijit Samanta 4 days, 16 hours ago

x^2-x-12
You can make any polynomial using these terms only by substituting values

Deep Kevadiya 4 days, 1 hour ago

Ragini Kumari 5 days, 6 hours ago

The zeroes of the polynomial are the values of X which satisfy the equation Y= F(X)
Zeroes of the polynomial are to be determined by how many times the line touches the x-axes .

Jeba Hussain 5 days, 19 hours ago

The zeros of polynomial are the values of x which satisfy the equation y = f(x)

Dhairya Sobti 5 days, 19 hours ago

Zeroes of the polynomial are those who can make a polynomial value zero

Abbhiijit Samanta 4 days, 16 hours ago

1

Dhairya Sobti 5 days, 18 hours ago

These values are deleted ☺️☺️☺️

Sameer Hsngrh 4 days, 16 hours ago

Hlo

Garvit Vats 5 days, 19 hours ago

3/4

Dhairya Sobti 5 days, 19 hours ago

Cos theta = 4/5 = B/H Therefore, Tan theta = P/B= 3/4 .

Harsh Gopal 6 days, 17 hours ago

Theta / 360⁰ × pi × r^2

Riya Kumari 5 days, 23 hours ago

x = -4 and y = -3

Prince Kumar Sah 6 days, 21 hours ago

√49 = √7×7 = 7

Ashish Majhi 6 days, 23 hours ago

7

Rahaf Rehas 1 week ago

7*7 = 49 So the √49 is 7

Naman Gupta 1 week, 1 day ago

7

Akashi Gupta 1 week, 1 day ago

√49 = 7
Triangle means a shape made up of three side or three straight line In triangle their are two types:- 1.sides. 2.angle SIDES:-. ANGLE:- ¹.equilateral ¹.acute ².isosceles ².obtuse ³.scalen. ³.right Similarities property:- RHS AAS SAS SSS A triangle are said to be similar when their sides are proportional and opposite angle are equal .

Akashi Gupta 1 week ago

Triangle

Deepti Mittal 6 days, 4 hours ago

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