An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15,000 ft., what is the distance (x) from the plane (P) to the observer (O)? (4 points) A right triangle is shown with angle O marked 42 degrees, hypotenuse marked x and the height marked 15000 feet. Select one: a. 10,035 feet b. 16,648 feet c. 20,188 feet d. 22,417 feet

D. 22,417 feet

Step-by-step explanation:

Fine the diagram in the attachment for proper elucidation. Using the SOH, CAH, TOA trigonometry identity to solve for the distance (x) from the plane (P) to the observer (O), the longest side x is the hypotenuse and the side facing the angle of elevation is the opposite.

Hypotenuse = x and Opposite = 15,000feet

According to SOH;

Hence the distance (x) from the plane P to the observer O is approximately 22,417 feet

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

the answer is a

step-by-step explanation:

39.00 x 24 = total installment price

carrying installment price - cash price

needed to save the .00/