The eyes of a basket ball player are 6 feet above the floor. He is at the free-throw line, which is 15 feet from the center of the basket rim. The center of the rim is 10 feet above the floor.
Before shooting the player decides to step back a few steps. His angle of elevation is now 12.5o. How far away from the center of the rim is he standing?
okay, the diagram shows the player and his distance with the center of the rim, for this I used the tan funtion =15/tan12.5, giving me 67.660, but that don't sound correct
+129852 Let's look at this....you're using the correct function.....note that the height of the basket doesnt change.....and we're looking for the adjacent side....note that this isn't 15 anymore, because the player has moved back from the FT line......
So....
tan 12.5 = 10/x
Note that x must now be greater than 15 .....can you take it from here??
Whoops...I made a mistake...I forgot that we said that the player's eyes are 6 ft from the floor....disregard the previous baloney.......
Note now that the basket is only 4ft feet above the player's line of sight..
So we have...
tan12.5 = 4/x
OK...I checked this answer...it makes sense........see if you can tell me what we get now for "x".......
+129852 Let's look at this....you're using the correct function.....note that the height of the basket doesnt change.....and we're looking for the adjacent side....note that this isn't 15 anymore, because the player has moved back from the FT line......
So....
tan 12.5 = 10/x
Note that x must now be greater than 15 .....can you take it from here??
Whoops...I made a mistake...I forgot that we said that the player's eyes are 6 ft from the floor....disregard the previous baloney.......
Note now that the basket is only 4ft feet above the player's line of sight..
So we have...
tan12.5 = 4/x
OK...I checked this answer...it makes sense........see if you can tell me what we get now for "x".......
yes, so if I was looking for the angle of elevation from his eye to the center of the rim, I would use arcsin? with 6 and 15?
+1006 Just take CPhill's ending equation and tweak it a bit.
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{12.5}}^\circ\right)} = {\frac{{\mathtt{4}}}{{\mathtt{x}}}}$$
To isolate 'x' is a bit tricky. First, you have to multiply the '4/x' by 'x' and do the same to the other side.
$${x}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{12.5}}^\circ\right)}\right)} = {\mathtt{4}}$$
Now divide both sides by tan(12.5)
$${\mathtt{x}} = {\frac{{\mathtt{4}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{12.5}}^\circ\right)}}}$$
Solve.
$${\frac{{\mathtt{4}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{12.5}}^\circ\right)}}} = {\mathtt{18.042\: \!834\: \!014\: \!643\: \!337\: \!4}}$$
The basketball player is now approx 18.04 feet from the center of the rim.
