Answer: -81
Step-by-step explanation:
Let x = the car moving towards west = 30
Let y = truck moving towards north = 40
dx/dt = -55
dy/dt = -60
We the use Pythagoras theorem to solve further where,
s² = x² + y²
s² = 30² + 40²
s² = 900 + 1600
s² = 2500
s =✓2500
s = 50
We then differentiate with respect to t, then we will have:
s² = x² + y²
2s ds/dt = 2x ds/dt + 2y dy/dt
50 ds/dt = (30 × -55) + (40 × -60)
50 ds/dt = -1650 - 2400
50 ds/dt = - 4050
ds/dt = -4050/50
ds/dt = -81
A communications tower is supported by two wires, connected at the same point on the ground. One is attached to the tower at D and the long-on at C. The angle AD makes with the ground is 30 ° and the angle between the two wires is 10 °. How much below the top of the tower is the shorter one attached?
Answer:
10.99 m
Step-by-step explanation:
✍️What we are basically asked to solve here is to find the distance between C and D.
To find CD, find the length of BC, and BD. Their difference will give us CD.
Thus, BC - BD = CD.
✍️Finding BC using trigonometric ratio formula:
[tex] \theta = 30 + 10 = 40 [/tex]
Opposite side = BC = ??
Adjacent side = 42 m
Thus:
[tex] tan(\theta) = \frac{opposite}{adjacent} [/tex]
Plug in the values
[tex] tan(40) = \frac{BC}{42} [/tex]
Multiply both sides by 42
[tex] tan(40) \times 42 = \frac{BC}{42} \times 42 [/tex]
[tex] 35.24 = BC [/tex]
BC = 35.24 m
✍️Finding BD using trigonometric ratio formula:
[tex] \theta = 30 [/tex]
Opposite side = BD = ??
Adjacent side = 42 m
Thus:
[tex] tan(\theta) = \frac{opposite}{adjacent} [/tex]
Plug in the values
[tex] tan(30) = \frac{BD}{42} [/tex]
Multiply both sides by 42
[tex] tan(30) \times 42 = \frac{BD}{42} \times 42 [/tex]
[tex] 24.25 = BD [/tex]
BD = 24.25 m
✍️How much below the top of the tower is the shorter one attached:
Thus,
BC - BD = CD
35.24 m - 24.25 m = 10.99 m