Given: circle k(O), m AM=125°, m EF=31°, m∠MAF=75°. Find: m∠AME
1 answer:
Answer:
58°
Step-by-step explanation:
If m AM=125°, then m∠AOM=125°.
If m∠MAF=75°, then m∠MOF=150° (because central angle MOF subtends on the same arc as inscribed angle MAF).
Thus,
m∠FOA=360°-150°-125°=85°.
If mEF=31°, then m∠EOF=31° (as central angle subtended on the arc EF).
Hence,
m∠EOA=m∠EOF+m∠FOA=31°+85°=116°.
Angle EOA is central angle subtended on arc EA, angle AME is inscribed angle subtended on arc AE, thus
m∠AME=1/2m∠EOA=58°.
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Answer:
6.667 ft/s
Step-by-step explanation:
Given
Height of street light = 15ft
Height of man = 6ft
Speed away from pole = 4ft/s
Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.
This forms a similar triangle (see attachment below).
From similar triangles, we have
(y - x)/6 = y/15 ----- Solve equation
15(y - x) = 6 * y
15y - 15x = 6y ---- Collect Like Terms
15y - 6y = 15x
9y = 15x --- divide through by 9
y = 15x/9
y = 5x/3 ---- differentiate with respect to time, t
dy/dt = 5/3 dx/dt
dx/dt represents rate of change of distance per time = 4ft/s
while dy/dt represents rate of movement of his shadow tips
dy/dt = 5/3 * 4
dy/dt = 20/3 = 6.667 ft/s
