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Every weekend me and my brother and my cousin go outside to play basketball and when i was the one who will shoot the ball to the ring i used the things i learned in math so I use SOH-CAH-TOA in measuring it. I want to measure the distance between me and the basketball ring. Supposed the height of the ring is 8ft from the ground and the distance between me and the basketball ring horizontally is 11.5 ft, and the angle of elevation from the ground is 50 degrees, now what is the distance of me and the ring?
Let x be the distance between me and the ring

need urgent answers with equation
by using SOH from SOHCAHTOA

Every Weekend Me And My Brother And My Cousin Go Outside To Play Basketball And When I Was The One Who Will Shoot The Ball To The Ring I Used The Things I Learn class=

Sagot :

Answer:

10 . 44 ft

Step-by-step explanation:

sin 50 = opposite leg / hypotenuse = 8 / x

8 / sin 50 = x    = 10 .44  ft

Tanweeleong195oxyqwpidn

Answer:

10.433cm

Step-by-step explanation:

Using SOH from TOACAHSOH,

[tex]sin50deg = \frac{opposite}{hypotenuse} [/tex]

From here we know there the opposite is 8ft (height of the ring) and hypotenuse is what we want to find, distance between you and the ring.

[tex]sin(50) = \frac{8}{x} \\ x \sin(50) = 8 \\ x = \frac{8}{ \sin(50) } \\ x = 10.443cm(rounded \: to \: 5sf)[/tex]

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