The angle of elevation of a jet plane from a point A on the ground is $60^{\circ}$. After a flight of 20 seconds at the speed of $432 km /$ hour, the angle of elevation changes to $30^{\circ} .$ If the jet plane is flying at a constant height, then its height is :

Question

If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where $ \alpha, \beta \in \mathbb{N} $, then the value of $ \alpha + \beta $ is ___________.

Answer 1

To determine the height of the jet plane flying at a constant altitude, we can solve the problem using trigonometric principles and the given information.

First, let's convert the speed of the jet plane from kilometers per hour to meters per second. Given the speed is 432 \text{ km/hr}, we convert it as follows: 432 \text{ km/hr} = \frac{432 \times 1000}{3600} \text{ m/s} = 120 \text{ m/s}.
The jet plane flies at a speed of 120 \text{ m/s} for 20 seconds, hence the distance traveled by the plane during this time is: 120 \times 20 = 2400 \text{ meters}.
We have two positions of the plane with different angles of elevation. Initially, the angle of elevation is 60^\circ, and finally, the angle of elevation becomes 30^\circ.
Let's denote the height of the plane as h, and use trigonometric relationships to find this height.
- From the initial position: \tan 60^\circ = \frac{h}{x}, where x is the distance of the initial position of the plane to point A on the ground. Since \tan 60^\circ = \sqrt{3}, we have: x = \frac{h}{\sqrt{3}}.
- From the final position: \tan 30^\circ = \frac{h}{x + 2400}. Since \tan 30^\circ = \frac{1}{\sqrt{3}}, we have: x + 2400 = h \cdot \sqrt{3}.
Now, let's set up the equations based on our findings: \begin{align*} \text{From first position:} & \quad x = \frac{h}{\sqrt{3}}, \\ \text{From second position:} & \quad x + 2400 = h \cdot \sqrt{3}. \end{align*}
We substitute the value of x from the first equation into the second equation: \frac{h}{\sqrt{3}} + 2400 = h \cdot \sqrt{3}.
Simplify and solve for h: \begin{align*} \frac{h}{\sqrt{3}} + 2400 &= h \cdot \sqrt{3}, \\ 2400 &= h \cdot \sqrt{3} - \frac{h}{\sqrt{3}}, \\ 2400 &= \frac{3h - h}{\sqrt{3}}, \\ 2400 &= \frac{2h}{\sqrt{3}}, \\ 2400 \times \sqrt{3} &= 2h, \\ h &= \frac{2400 \sqrt{3}}{2}, \\ h &= 1200 \sqrt{3} \text{ meters}. \end{align*}

Therefore, the height of the jet plane is 1200 \sqrt{3} \text{ meters}.

Answer 2

To determine the height of the jet plane flying at a constant altitude, we can solve the problem using trigonometric principles and the given information.

First, let's convert the speed of the jet plane from kilometers per hour to meters per second. Given the speed is 432 \text{ km/hr}, we convert it as follows: 432 \text{ km/hr} = \frac{432 \times 1000}{3600} \text{ m/s} = 120 \text{ m/s}.
The jet plane flies at a speed of 120 \text{ m/s} for 20 seconds, hence the distance traveled by the plane during this time is: 120 \times 20 = 2400 \text{ meters}.
We have two positions of the plane with different angles of elevation. Initially, the angle of elevation is 60^\circ, and finally, the angle of elevation becomes 30^\circ.
Let's denote the height of the plane as h, and use trigonometric relationships to find this height.
- From the initial position: \tan 60^\circ = \frac{h}{x}, where x is the distance of the initial position of the plane to point A on the ground. Since \tan 60^\circ = \sqrt{3}, we have: x = \frac{h}{\sqrt{3}}.
- From the final position: \tan 30^\circ = \frac{h}{x + 2400}. Since \tan 30^\circ = \frac{1}{\sqrt{3}}, we have: x + 2400 = h \cdot \sqrt{3}.
Now, let's set up the equations based on our findings: \begin{align*} \text{From first position:} & \quad x = \frac{h}{\sqrt{3}}, \\ \text{From second position:} & \quad x + 2400 = h \cdot \sqrt{3}. \end{align*}
We substitute the value of x from the first equation into the second equation: \frac{h}{\sqrt{3}} + 2400 = h \cdot \sqrt{3}.
Simplify and solve for h: \begin{align*} \frac{h}{\sqrt{3}} + 2400 &= h \cdot \sqrt{3}, \\ 2400 &= h \cdot \sqrt{3} - \frac{h}{\sqrt{3}}, \\ 2400 &= \frac{3h - h}{\sqrt{3}}, \\ 2400 &= \frac{2h}{\sqrt{3}}, \\ 2400 \times \sqrt{3} &= 2h, \\ h &= \frac{2400 \sqrt{3}}{2}, \\ h &= 1200 \sqrt{3} \text{ meters}. \end{align*}

Therefore, the height of the jet plane is 1200 \sqrt{3} \text{ meters}.

The angle of elevation of a jet plane from a point A on the ground is $60^{\circ}$. After a flight of 20 seconds at the speed of $432 km /$ hour, the angle of elevation changes to $30^{\circ} .$ If the jet plane is flying at a constant height, then its height is :

The Correct Option is D

Solution and Explanation

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