Given: Lines AB and CD intersects at O such that ∠BOD = 40° and ∠AOC + ∠BOE = 70° ...(1) ∠AOC = ∠BOD [∵ Vertically Opposite Angles] Hence, ∠AOC = 40° [∵ ∠BOD = 40°] Therefore, from the equation (1), we have, 40° + ∠BOE = 70° ⇒ ∠BOE = 70° - 40° = 30° Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straigRead more
Given: Lines AB and CD intersects at O such that ∠BOD = 40° and
∠AOC + ∠BOE = 70° …(1)
∠AOC = ∠BOD [∵ Vertically Opposite Angles]
Hence, ∠AOC = 40° [∵ ∠BOD = 40°]
Therefore, from the equation (1), we have,
40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30°
Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straight line]
⇒ 70° + ∠COE = 180° [From the equation (1)]
⇒ ∠COE = 180° – 70° = 110° and
Relex ∠COE = 360° – ∠COE = 360° – 110° = 250°
(iv) 3 = 2x +y ⇒ y = 3 - 2x Putting x = 0, we have, y = 3 - 2 × 0 = 3 Putting x = 1, we have, y = 3 - 2 × 1 = 1 Hence, G(0,3) and H(1,1) are the solutions of the equation.
(iv) 3 = 2x +y
⇒ y = 3 – 2x
Putting x = 0, we have, y = 3 – 2 × 0 = 3
Putting x = 1, we have, y = 3 – 2 × 1 = 1
Hence, G(0,3) and H(1,1) are the solutions of the equation.
Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x - y = 2. There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.
Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x – y = 2.
There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.
In Fig. 6.13, lines AB and CD intersect at O.
Given: Lines AB and CD intersects at O such that ∠BOD = 40° and ∠AOC + ∠BOE = 70° ...(1) ∠AOC = ∠BOD [∵ Vertically Opposite Angles] Hence, ∠AOC = 40° [∵ ∠BOD = 40°] Therefore, from the equation (1), we have, 40° + ∠BOE = 70° ⇒ ∠BOE = 70° - 40° = 30° Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straigRead more
Given: Lines AB and CD intersects at O such that ∠BOD = 40° and
See less∠AOC + ∠BOE = 70° …(1)
∠AOC = ∠BOD [∵ Vertically Opposite Angles]
Hence, ∠AOC = 40° [∵ ∠BOD = 40°]
Therefore, from the equation (1), we have,
40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30°
Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straight line]
⇒ 70° + ∠COE = 180° [From the equation (1)]
⇒ ∠COE = 180° – 70° = 110° and
Relex ∠COE = 360° – ∠COE = 360° – 110° = 250°
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Axiom 5: The whole is greater than the part. Since this is true for anything in any part of the world, this is a universal truth.
Axiom 5:
See lessThe whole is greater than the part.
Since this is true for anything in any part of the world, this is a universal truth.
Draw the graph of each of the following linear equations in two variables: 3 = 2x + y
(iv) 3 = 2x +y ⇒ y = 3 - 2x Putting x = 0, we have, y = 3 - 2 × 0 = 3 Putting x = 1, we have, y = 3 - 2 × 1 = 1 Hence, G(0,3) and H(1,1) are the solutions of the equation.
(iv) 3 = 2x +y
See less⇒ y = 3 – 2x
Putting x = 0, we have, y = 3 – 2 × 0 = 3
Putting x = 1, we have, y = 3 – 2 × 1 = 1
Hence, G(0,3) and H(1,1) are the solutions of the equation.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x - y = 2. There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.
Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x – y = 2.
See lessThere are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Given equation of line: 3y = ax + 7. Putting x = 3 and y = 4, we have, 3 x 4 = a x3 +7 ⇒ 12 = 3a + 7 ⇒ 12 - 7 = 3a ⇒ a = 5/3
Given equation of line: 3y = ax + 7.
See lessPutting x = 3 and y = 4, we have, 3 x 4 = a x3 +7
⇒ 12 = 3a + 7 ⇒ 12 – 7 = 3a
⇒ a = 5/3