Question

Can someone please help me with this geometry question I will mark brainliest

Answer 1

Answer:

hypotenuse = 8$\sqrt{3}$ , leg = 12

Step-by-step explanation:

Using the cosine and tangent ratios in the right triangle and the exact values

cos60° = $\frac{1}{2}$ , tan60° = $\sqrt{3}$ , then

cos60° = $\frac{adjacent}{hypotenuse}$ = $\frac{4\sqrt{3} }{hypotenuse}$ = $\frac{1}{2}$ ( cross- multiply )

hypotenuse = 4$\sqrt{2}$ × 2 = 8$\sqrt{2}$

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tan60° = $\frac{opposite}{adjacent}$ = $\frac{opposite}{4\sqrt{3} }$ = $\sqrt{3}$ ( multiply both sides by 4$\sqrt{3}$ )

opposite = 4$\sqrt{3}$ × $\sqrt{3}$ = 12

Answer 2

Answer:

Lengths are 8√3 and 12

Step-by-step explanation:

» From trigonometric ratios, using our angle as 60°

${ \tt \cos( \theta) = \frac{adjacent}{hypotenuse} } \\ \\ { \tt{ \cos( 60 \degree) = \frac{4 \sqrt{3} }{hypotenuse} }} \\ \\ { \tt{hypotenuse = \frac{4 \sqrt{3} }{ \cos(60 \degree) } }} \\ \\ { \tt{hypotenuse = \frac{4 \sqrt{3} }{0.5} }} \\ \\ { \boxed{ \tt{hypotenuse = 8 \sqrt{3} }}}$

» Using 30° as our angle:

${ \tt{ \tan( \theta) = \frac{opposite}{adjacent} }} \\ \\ { \tt{ \tan(30 \degree) = \frac{4 \sqrt{3} }{adjacent} }} \\ \\ { \tt{adjacent = \frac{4 \sqrt{3} }{ \frac{1}{ \sqrt{3} } } }} \\ \\ { \tt{adjacent = \frac{(4 \sqrt{3} ) \times ( \sqrt{3}) }{1} }} \\ \\ { \tt{adjacent = 12}}$