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KiRa [710]
3 years ago
6

Please help me thank you so much

Mathematics
1 answer:
Romashka [77]3 years ago
4 0

Answer:

<u>Solving</u><u> </u><u>for</u><u> </u><u>d</u><u> </u><u>:</u>

» We're gonna use the sine trigonometric ratios angle 60°

{ \tt{ \sin( \theta) =  \frac{opposite}{hypotenuse}  }} \\  \\ { \tt{ \sin(60 \degree)  =  \frac{d}{2} }} \\  \\ { \tt{d = 2 \times  \sin(60 \degree) }} \\ { \tt{d = 2 \times  \frac{ \sqrt{3} }{2} }} \\   \\ { \boxed{ \tt{ \: d =  \sqrt{3} } \: }}

<u>Solving</u><u> </u><u>for</u><u> </u><u>b</u><u> </u><u>:</u>

» We're gonna use cosine trigonometric ratios of angle 45°

{ \tt{ \cos( \theta) =  \frac{adjacent}{hypotenuse}  }} \\  \\ { \tt{ \cos(45 \degree)  =  \frac{b}{4} }} \\  \\ { \tt{b = 4 \times  \cos(45 \degree) }} \\ { \tt{b = 4 \times  \frac{1}{ \sqrt{2}  }  =  \frac{4}{ \sqrt{2} } }} \\  \\  { \tt{b = ( \frac{4}{ \sqrt{2} }) \times  \frac{ \sqrt{2} }{ \sqrt{2} }  }} \\  \\ { \tt{b =  \frac{4 \sqrt{2} }{2} }} \\  \\ { \boxed{ \tt{ \: b = 2 \sqrt{2}  \: }}}

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