A searchlight located 200 m from a weather office is shined directly overhead. If the angle of elevation to the spot of light on the clouds is 35 degrees, how high is the cloud ceiling?
i used the sin function
+130511 I'm assuming that the "directly overhead" refers to "directly ovehead" the weather office.
We need the tangent function, here. Note that, we know the amgle of elevation, and the distance that the searchlight is from the weather office. And we're looking for the opposite side of the angle (the height of the clouds above the ground).....so we have
tan(35) = h/200 where h is the height of the clouds...
I can provide a general diagram....if you need it......
+130511 I'm assuming that the "directly overhead" refers to "directly ovehead" the weather office.
We need the tangent function, here. Note that, we know the amgle of elevation, and the distance that the searchlight is from the weather office. And we're looking for the opposite side of the angle (the height of the clouds above the ground).....so we have
tan(35) = h/200 where h is the height of the clouds...
I can provide a general diagram....if you need it......
