Question

An airplane left airport A flying on a course of 72 degrees.

Answer 1

Answer:

Distance for which Aeroplane can be in contact with Airport B is = 396.34 km

Step-by-step explanation:

In the question,

We have an Airport at point A and another at point B.

Now,

Airplane flying at the angle of 72° with vertical catches signals from point D.

Distance travelled by Airplane, AD = 495 km

Now, Let us say,

AB = x

So,

In triangle ABD, Using Cosine Rule, we get,

$cos(90-72) =cos18= \frac{AB^{2}+AD^{2}-BD^{2}}{2.AD.AB}$

So,

On putting the values, we get,

$cos18 = \frac{x^{2}+495^{2}-300^{2}}{2(495)(x)}\\0.951(990x)=x^{2}+245025-90000\\x^{2}-941.54x+155025=0$

Therefore, x is given by,

x = 212.696, 728.844

So,

The value of x can not be 212.696 as the length of LB (radius) itself is 300 km.

So,

x = 728.844 km

So,

AL = AB - BL

AL = x - 300

AL = 728.844 - 300

AL = 428.844 km

Now, in the circle from a property of secants we can say that,

AL x AM = AD x AC

So,

428.844 x (728.844 + 300) = 495 x AC

441213.576 = 495 x AC

AC = 891.34 km

So,

The value of CD is given by,

CD = AC - AD

CD = 891.34 - 495

CD = 396.34 km

Therefore, the distance for which the Aeroplane can still be in the contact with Airport B is 396.34 km.