Question #41c0d

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Question #41c0d

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Answer 1 · 2016-11-28T18:48:00

$tan x = \frac{3}{4}$ $tanx=3/4$

Explanation:

Draw the angle in the first quadrant. The hypotenuse will be $5$ $5$ and the adjacent will be $4$ $4$ , since

$cos x = \frac{adjacent}{hypotenuse}$ $cosx = "adjacent"/"hypotenuse"$

Then use the Pythagorean Theorem to solve for the remaining side, which comes out to 3.

$a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$

$4^{2} + b^{2} = 5^{2}$ $4^2 + b^2 = 5^2$

$16 + b^{2} = 25$ $16 + b^2 = 25$

$b^{2} = 25 - 16$ $b^2 = 25 - 16$

$b^{2} = 9$ $b^2 = 9$

$b = \sqrt{9}$ $b = sqrt 9$

$b = 3$ $b = 3$

So the side opposite to the angle $x$ $x$ in the triangle is $3$ $3$ . Then, since

$tan (x) = \frac{opposite}{adjacent}$ $tan(x) = "opposite"/ "adjacent"$

you have

$tan x = \frac{3}{4}$ $tanx= 3/4$

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Question #41c0d

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