Two airplanes are approaching the airport tower at the same altitude. radar shows that plane a is 20 miles due east of the tower and that plane b is 22 miles at 15° north of east. (see the drawing provided above).

how far apart are the two planes? round to the nearest hundredths.

miles

40 miles

Step-by-step explanation:

We can solve this by drawing a triangle if one plane is east of the tower and the other north east. We assume that the tower is the other point of the triangle. We have an angle of 15 degrees connecting tower to plan B. We can sine rule to solve this problem.

The sine rule is:

sinA/a=sinB/b=sinC/c

As we know one angle and two sides we can solve for the one side and two angles as we have 3 unknowns and 3 known values:

Let plane A have angle A and side a, plane B have angle B and side b:

sinA/a=sinB/b

sinA/20=sin15/22

sinA

A = sin^-1 (0.235)=13.6 degrees

Therefore angle C is:

We can solve for c:

sinB/b=sinC/c

The planes are 40.6 miles apart

dd

step-by-step explanation:

dd

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)