\(\frac{6}{5}\)
\(\frac{9}{5}\)
\(\frac{4}{3}\)
\(\frac{7}{3}\)
Apply Pythagoras Theorem in NAB,
\(NA = \sqrt{15^2 - 9^2}\)
\(NA=12\)
\(\frac{h}{15} = \tan \theta = \frac{2}{3}\)
\(h = 10\ units\)
\(\cot \alpha = \frac{12}{10}\)
\(\cot \alpha = \frac{6}{5}\)
So, the correct option is (A): \(\frac{6}{5}\)
If the origin is shifted to a point \( P \) by the translation of axes to remove the \( y \)-term from the equation \( x^2 - y^2 + 2y - 1 = 0 \), then the transformed equation of it is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).
The length of the perpendicular drawn from the point to the line is the distance of a point from a line. The shortest difference between a point and a line is the distance between them. To move a point on the line it measures the minimum distance or length required.
The following steps can be used to calculate the distance between two points using the given coordinates:
Note: If the two points are in a 3D plane, we can use the 3D distance formula, d = √(m2 - m1)2 + (n2 - n1)2 + (o2 - o1)2.
Read More: Distance Formula