Answer:
BD = <u>1</u> unit
AD = <u>1</u> unit
AB = <u>1.6</u> units
AC = <u>1.6</u> units
Step-by-step explanation:
In the picture attached, the triangle ABC is shown.
Given that triangle ABC is isosceles, then ∠B = ∠C
∠A + ∠B + ∠C = 180°
36° + 2∠B = 180°
∠B = (180° - 36°)/2
∠B = ∠C = 72°
From Law of Sines:
sin(∠A)/BC = sin(∠B)/AC = sin(∠C)/AB
(Remember that BC is 1 unit long)
AB = AC = sin(72°)/sin(36°) = 1.6
In triangle ABD, ∠B = 72°/2 = 36°, then:
∠A + ∠B + ∠D = 180°
36° + 36° + ∠D = 180°
∠D = 180° - 36° - 36° = 108°
From Law of Sines:
sin(∠A)/BD = sin(∠B)/AD = sin(∠D)/AB
(now ∠A = ∠B)
BD = AD = sin(∠A)*AB/sin(∠D)
BD = AD = sin(36°)*1.6/sin(108°) = 1
Answer:
1176
Step-by-step explanation:
12 times 7 equals 84.84 times 14 is 1176
Answer:
The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.
Step-by-step explanation:
Given information:
A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.
We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.
According to Pythagoras
.... (1)
Put z=1 and y=2, to find the value of x.
Taking square root both sides.
Differentiate equation (1) with respect to t.
Divide both sides by 2.
Put 
, y=2, in the above equation.
Divide both sides by 2.
Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.
(Add5) 1/9..(add 7) 1/16... (add9) 1/25... (add 11) and get -> 1/36
Answer:
c and d
Step-by-step explanation:
i think
