Trig Function question (1 Viewer)

[pi-ka-chew]

the moon subtends an angle of 31' at an observation point on Earth 400 000 km away. Use the fact that the diameter of the moon is approximately equal to an arc of a circle whose centre is the point of observation to show that the diameter of the moon is approximately 3600 km. Please help. thanks in advance

 

[life92]

Okay, since we know that the diameter of the moon is approximately the length of the arc with the centre as the observation point,
we use the formula l = r theta, where l is the arc length, r is the radius and theta is the angle subtended at the centre

Now the radius = 400 000 km

theta must be in radians so...
31' = 31/60 degrees
31/60 degrees = 31/60 * pi/180 rads

Therefore, the diameter of the moon = 400 000 * 31/60 * pi/180
Plug that into a calculator and you get around ~3605km, which is the required answer, as it asks for approximately 3600km.

Hope that helps !

 

[pi-ka-chew]

ah...i see now. i didn't know they were the radius and the angle subtended.
thanks a lot.