An airplane is flying at an elevation of 5150 ft, directly above a straight highway. two motorists are driving cars on the highway on opposite sides of the plane, and the angle of depression to one car is 34° and to the other is 54°. how far apart are the cars? (round your answer to the nearest foot.)

The cars are 10702ft apart

Step-by-step explanation:

Height of the plane from the high way = 5150ft

Considering Triangle ABD, we will be using SOHCAHTOA

Tan = Opposite/ Adjacent

Tan 54° = Y/5150

Y = 5150*tan54°

Y = 7088 ft

Considering Triangle ADC, Tan = Opposite/ Adjacent is also applicable

Tan 34° = X/5150

X = 5150*tan 54°

X = 3614ft

The distance between the two cars = (X+Y)

= 7088 + 3614

= 10702ft

found it xd it ok yt

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180