Question

Solve the question along with diagram.​

Answer 1

$\huge{\bold{\underline{\underline{Solution:-}}}}$

$\large{\bold{\underline{To Find}}}$

Speed of plane (v) = ?.

$\large{\bold{\underline{\underline{Step - Step -by - Explanation}}}}$

${ED = AB = 1500 \sqrt{3} meters}$

${ In \: \triangle \: ABC}$

${ \tan{60} = \frac{AB}{BC}}$

${ \sqrt{3} = \frac{1500 \sqrt{3}}{BC}}$

BC = 1500 meters.

${ In \: \triangle \: ABC}$

${ \tan{30} = \frac{ED}{CD}}$

${ \frac{1}{ \sqrt{3}} = \frac{1500 \sqrt{3}}{CD}}$

CD = 4500 meters.

BD = 4500 - 1500 = 3000 metres.

${Speed = \frac{Distance}{time}}$

${Speed = \frac{3000}{15}}$

Speed = 200 m/s.

$\huge{\boxed{\boxed{ v = 200 m/s}}}$

so, the speed is 200m/s.

Answer 2

Answer :

The speed of the plane is 200m/s.

Step-by-step explanation :

Given Information -

The angle of Jet plane from point A on the ground is 60°

After the flight of 15 seconds, The angle of elevation changes into 30°

The jet plane is still flying at a constant height of 1500 √3.

To Find :

The Speed of the Jet Plane.

Solution :

Consider the -

Point of observation as - A.

Two points of the plane as - C , E.

According to the Given Information,

⟶ ∠BAC = 60°

⟶ ∠DAE = 30°

⟶ CB = 1500 √3.

Since,

The jet plane is still flying at a constant height of 1500 √3.

∴ CB = DE 1500 √3.

In the right triangle ABC,

We have,

$\sf\\ \\\implies Tan\;60\textdegree=\dfrac{BC}{AB}\\ \\ \\ \implies \sqrt{3} = \dfrac{1500\;\sqrt{3}}{AB}\\ \\ \\\implies AB = \dfrac{1500\;\sqrt{3}}{\sqrt{3}}\\ \\ \\\implies AB =1500m.\quad\quad...1$

In right triangle ADE,

We have,

$\sf \\ \\\implies Tan\;30\textdegree = \dfrac{DE}{AD}\\ \\ \\\implies \dfrac{1}{\sqrt3} = \dfrac{DE}{AB+BD}\\ \\ \\\implies \dfrac{1}{\sqrt3} = \dfrac{1500\;\sqrt{3}}{AB+BD}\\ \\ \\\implies AB+BD = 1500\;\sqrt{3} \times \sqrt{3}\\ \\ \\\implies AB+BD=4500.\quad\quad...2$

Now,

Put the values of 1 in 2,

$\sf\\ \\\implies 1500+BD = 4500.\\ \\ \\\implies BD = 3000$

∵ Distance travelled in 15 seconds.

⇒ CE = BD = 3000 meters.

Now,

We know that :

$\bigstar\;\boxed{\sf Speed = \dfrac{Distance}{Time}}}$

Values in Equation,

Speed of the Plane,

$\sf\\ \\\implies \dfrac{3000}{15}\\ \\ \\ \implies Speed\;of\;the\;plane = 200\;m\s.$

∴ The speed of the plane is 200m/s.