Maths

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is

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Contributor-Level 10

Let base = b

$t a n 60 ° = \frac{h}{b}$

$t a n 30 ° = \frac{h ? 20}{b}$

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New question posted

a year ago

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is

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New answer posted

a year ago

Maths Maths Statistics

The number of values of $a \in N$ such that the variance of 3, 7, 12, a, 43 a is a natural number is

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Maths Relations and Functions

Contributor-Level 10

$M e a n = \frac{3 + 12 + 7 + a + (43 - a)}{5} = 13$

Variance = $\frac{3^{2} + 1 2^{2} + 7^{2} + a^{2} + {(43 - a)}^{2}}{5} - {(13)}^{2}$

$\Rightarrow \frac{2 a^{2} - a + 1}{5} \to N a t u r a l n u m b e r$

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4 λ o r 4 λ + 1 f r o m .$

As each square form is $4 λ o r 4 λ + 1$

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a year ago

The probability that a relation R from ${x, y} t o {x, y}$ is both symmetric and transitive, is equal to

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Contributor-Level 10

Total number of possible relation = $2^{n^{2}} = 2^{4} = 16$

Favourable relations = $?, {(x, x)}, {(y, y)}$

${(x, x), (y, y)}$

${(x, x), (y, y), (x, y), (y, x)}$

Probability = $\frac{5}{16}$

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New answer posted

a year ago

Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is

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Contributor-Level 10

Mid point of BC is $\frac{1}{2} (5 \hat{i} + (α - 2) \hat{j} + 9 \hat{k})$

$\bar{A B} = \hat{i} + (α - 4) \hat{j} + \hat{k}$

$\bar{A C} = \hat{i} + (- 2 - α) \hat{j} + \hat{k}$

For = 1, $\bar{A B}$ and $\bar{A C}$ will be collinear. So for non collinearity

= 2

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New answer posted

a year ago

Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y +2z = 16. Let T be a plane passing through the point Q and contains the line $\vec{r} = - \hat{k} + λ (\hat{i} + \hat{j} + 2 \hat{k}), λ \in R .$ Then, which of the following points lies on T?

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$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 1}{2} = \frac{- 2 (1 + 4 + 2 - 16)}{1 + 2^{2} + 2^{2}}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$| \begin{matrix} x & y & z + 1 \\ 3 & 6 & 6 \\ 1 & 2 \end{matrix} | = 0$

2x – z = 1

option (B) satisfies.

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a year ago

The distance of the origin from the centroid of the triangle whose two sides have the equations x – 2y + 1 = 0 and 2x – y – 1 = 0 and whose orthocenter is $(\frac{7}{3}, \frac{7}{3}) i s :$

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Contributor-Level 10

Let AB $\equiv x - 2 y + 1 = 0$

AC $\equiv 2 x - y + 1 = 0$

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

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New answer posted

a year ago

Let $\frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z + 3}{- 1}$ lie on the plane px – qy + z = 5, for some $p, q \in R .$ The shortest distance of the plane form the origin is

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Contributor-Level 10

Line $⊥$ to the normal

⇒ 3p + 2q – 1 = 0

$(2, - 1, - 3)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin = $| \frac{5}{\sqrt[]{1 5^{2} + {(- 22)}^{2} + 1^{2}}} |$ $=$ √5/142

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New answer posted

a year ago

Let a triangle ABC be inscribed in the circle $x^{2} - \sqrt[]{2} (x + y) + y^{2} = 0$ such that $∠ B A C = \frac{π}{2} .$ If the length of side AB is $\sqrt[]{2},$ then the area of the $Δ A B C$ is equal to

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Contributor-Level 10

${(x - \frac{1}{\sqrt[]{2}})}^{2} + {(y - \frac{1}{\sqrt[]{2}})}^{2} = 1$

here AB = $\sqrt[]{2}$ , BC = 2, AC = $\sqrt[]{2}$

area = $\frac{1}{2} * \sqrt[]{2} * \sqrt[]{2} = 1$

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New answer posted

a year ago

Let P : y² = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is

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