The measure of ∠ACB will be 110°
<u><em>Explanation</em></u>
According to the diagram below, 
and 
are the perpendicular bisectors of 
and 
respectively and they intersect side 
at points 
and 
respectively.
So, 
and 
Now, <u>according to the 
postulate</u>, ΔAPE and ΔCPE are congruent each other. Also, ΔCFQ and ΔBFQ are congruent to each other.
That means, ∠PCE = ∠PAE and ∠FCQ = ∠FBQ
As ∠CPQ = 78° , so ∠PCE + ∠PAE = 78° or, ∠PCE = 
° and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = 
°
Now, in triangle CPQ, ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°
Thus, ∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°
Answer:
Plane B is closer.
Step-by-step explanation:
Plane B is closer.
The distance for Plane A is 7.9 km.
The distance for Plane B is 7.4 km.
Step-by-step explanation:
To find the distance to the tower, altitude must be converted to kilometers. Each 1000 ft is 0.3048 km, so the heights of the planes are ...
-- Plane A: (20 thousand ft)*(0.3048 km/thousand ft) = 6.096 km
-- Plane B: (8 thousand ft)*(0.3048 km/thousand ft) = 2.4384 km
The Pythagorean theorem can be used to find the distance from each plane to the tower:
-- distance = √((ground distance)² + (height)²)
-- Plane A distance = √(5² +6.096²) ≈ √62.16 ≈ 7.9 . . . km
-- Plane B distance = √(7² +2.4384²) ≈ √54.95 ≈ 7.4 . . . km
The distance for Plane B is shorter, so Plane B is closer to the tower.
Sorry if it's low quality and blurry.
