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  • 2 answers

Naba Meher 4 years, 1 month ago

You should find the distance between three pairs of vertices AB,BC,CA

Jatin Kushwaha 4 years, 1 month 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
  • 5 answers

Abhishek Pandey 4 years, 1 month ago

No solution

Sania Mirza 4 years, 1 month ago

Thanks

Chahak Dalmia 4 years, 1 month ago

no solution because there are given 2 values of y so both the line of y are Parallel to each other then they have no solution

Navya Sri Mekala 4 years, 1 month ago

One solution

Sumit Yadav 4 years, 1 month ago

No solution
  • 1 answers

Nitish Kumar 4 years, 1 month ago

From LHS. From RHS =(3+1) (3-√3). =3√3-2√3/2 = 3√3+3-3-√3. =6√3-2√3/2 =2√3. =4√3/2 =2√3 LHS. = RHS.
  • 1 answers

Sumit Yadav 4 years, 1 month 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)
  • 1 answers

Sumit Yadav 4 years, 1 month 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
  • 3 answers

Aseem Mahajan 4 years, 1 month 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]

Aseem Mahajan 4 years, 1 month 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:

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

Prajan Elango 4 years, 1 month ago

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

🤟Royal Thakur 🤟 4 years, 1 month ago

??Wow bro ??????

Aseem Mahajan 4 years, 1 month ago

Edited table :

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

Aseem Mahajan 2 years, 4 months 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

More information:

[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]

Aseem Mahajan 4 years, 1 month ago

17
  • 3 answers

Aseem Mahajan 4 years, 1 month ago

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

Its Nav Sandhu ✌✌ 4 years, 1 month ago

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

Aseem Mahajan 4 years, 1 month ago

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

Bhavya Maheshwari 4 years, 1 month 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 4 years, 1 month ago

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

Ansh Tyagi 4 years, 1 month ago

45× 2 = 90

Sakshi Kumari 4 years, 1 month ago

45×2 = 90

X-O Harsh 4 years, 1 month ago

(45)^2=90

Aastha Lilhore 4 years, 1 month ago

45*2 = 90
  • 2 answers

Its Nav Sandhu ✌✌ 4 years, 1 month ago

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

Prateek Aggarwal 4 years, 1 month ago

D 2520
  • 1 answers

Monarch Pathak¹³ 4 years, 1 month 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
  • 2 answers

Subodh Kumar 4 years, 1 month ago

Hi

Rohan Kandulna 4 years, 1 month ago

Questions no 1
  • 1 answers

Rohan Kandulna 4 years, 1 month ago

Experience 7.1
  • 0 answers
  • 2 answers

Abhijit Samanta 4 years, 1 month ago

x^2-x-12

Sahil Vinod Darekar 4 years, 1 month ago

You can make any polynomial using these terms only by substituting values
  • 4 answers

Ragini Kumari 4 years, 1 month ago

The zeroes of the polynomial are the values of X which satisfy the equation Y= F(X)

Kaif Khan 4 years, 1 month ago

Zeroes of the polynomial are to be determined by how many times the line touches the x-axes .

Jeba Hussain 4 years, 1 month ago

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

Chirayu Sobti 4 years, 1 month ago

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

Abhijit Samanta 4 years, 1 month ago

1

Chirayu Sobti 4 years, 1 month ago

These values are deleted ☺️☺️☺️
  • 3 answers

Sameer Hsngrh 4 years, 1 month ago

Hlo

Garvit Vats 4 years, 1 month ago

3/4

Chirayu Sobti 4 years, 1 month ago

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

Harsh Gopal 4 years, 1 month ago

Theta / 360⁰ × pi × r^2
  • 1 answers

Riya Kumari 4 years, 1 month ago

x = -4 and y = -3
  • 0 answers
  • 5 answers

Prince Kumar Sah 4 years, 1 month ago

√49 = √7×7 = 7

Ashish Majhi 4 years, 1 month ago

7

Rahaf Rehas 4 years, 1 month ago

7*7 = 49 So the √49 is 7

Naman Gupta 4 years, 1 month ago

7

Ayushi Gupta 4 years, 1 month ago

√49 = 7
  • 1 answers

Kaif Khan 4 years, 1 month ago

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 .
  • 1 answers

Ayushi Gupta 4 years, 1 month ago

Triangle

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