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PLEASE EXPLAIN!

 Feb 6, 2024

Best Answer 

 #1
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By pythagorean theorem, with diagonal EF, we know triangle EFG has side length EG = \sqrt{64 - 36} = 2\sqrt{7}

Now you can do trig: secE = 1/cosE. cosE = adj/hyp, so secE = hyp/adj. hypotenuse = 8, adjacent = 2\sqrt{7}

So, secE = 8/2\sqrt{7} = 4/\sqrt{7} = \(4\sqrt{7}/7\) after rationalizing the denominator.

cosF = adjacent/hypotenuse = 6/8 = 3/4

tanF = sinF/cosF (remember this). We know cosF is 3/4, and for sinF = opp/hyp, where opposite = 2\sqrt{7}, hypotenuse = 8.

2\sqrt{7}/8 = \sqrt{7}/4 thus, sinF/cosF = (\sqrt{7}/4)/(3/4) = \(\sqrt{7}/3\)

 Feb 6, 2024
edited by proyaop  Feb 6, 2024
 #1
avatar+1622 
+1
Best Answer

By pythagorean theorem, with diagonal EF, we know triangle EFG has side length EG = \sqrt{64 - 36} = 2\sqrt{7}

Now you can do trig: secE = 1/cosE. cosE = adj/hyp, so secE = hyp/adj. hypotenuse = 8, adjacent = 2\sqrt{7}

So, secE = 8/2\sqrt{7} = 4/\sqrt{7} = \(4\sqrt{7}/7\) after rationalizing the denominator.

cosF = adjacent/hypotenuse = 6/8 = 3/4

tanF = sinF/cosF (remember this). We know cosF is 3/4, and for sinF = opp/hyp, where opposite = 2\sqrt{7}, hypotenuse = 8.

2\sqrt{7}/8 = \sqrt{7}/4 thus, sinF/cosF = (\sqrt{7}/4)/(3/4) = \(\sqrt{7}/3\)

proyaop Feb 6, 2024
edited by proyaop  Feb 6, 2024

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