Can you help me with these math problem ?10 points
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Can you help me with these math problem ?10 points

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
cos(75°) = (adjacent/hypotenuse) = L/50, so L = 50·cos(75°) = 12.9 ft (approximately).-I couldnt get the image to load either (my firewall was in the way, I think). Now that I have seen it (at a different computer),......
If the distance from the top of a building to the tip of its shadow is 50 feet and the sun makes a 75 degree angle with the wall as shown in the image below, what is the length of its shadow?

http://curriculum.kcdistancelearning.com…

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50 = the hypotenuse of a right triangle whose shorter leg is the length (L) of the shadow.
cos(75°) = (adjacent/hypotenuse) = L/50, so L = 50·cos(75°) = 12.9 ft (approximately).

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I couldn't get the image to load either (my firewall was in the way, I think). Now that I have seen it (at a different computer), the answer is 50·sin(75°) = 48.3 ft.

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i may be mistaken since but I couldn't get the image to load...but i think you can use the 75 deg, as the angle if the shadow to the wall using the Z-rule (or w.e it's called). therefore, you can make a right angle triangle with the wall and shadow, with the hypotenuse being 50ft and the other sides unknown.

with x being the length of the shadow,

length of the shadow: cos75=x/50

x=50cos75 = 12.9 ft?
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