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VLD [36.1K]
2 years ago
6

Please help solve! I tried but I’m messing this up, would really appreciate it!

Mathematics
1 answer:
STALIN [3.7K]2 years ago
7 0

Answer:

<em>K = </em>\frac{4r}{n} -a<em> n </em>\neq 0<em />

Step-by-step explanation:

We have our equation r = \frac{n}{4} (a+K) and are looking for K

First we have to switch sides :

\frac{n}{4} (a+K) = r

Since we have a denominator of 4, we multiply both sides by 4 :

n (a+K) = 4r

We then divide both sides by n :

a+K = \frac{4r}{n} ; n\neq 0

Subtract a from both sides :

K = \frac{4r}{n} -a; n\neq 0

^^^

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Given:

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Using trigonometric ratio formula:

$\sin\theta =\frac{\text{opposite }}{\text{hypotenuse}}

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$\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}

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$\tan \theta=\frac{\text { opposite side }}{\text { adjacent side }}

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