Question

PLEASE HELPPP!!!!!!!

calculate tan0

Answer 1

use the pythagorean theorem to find the missing leg/side \(\bf c^2 = a^2 + b^2\implies \sqrt{c^2 -a^2} = b\)

once you find the missing leg, keep in mind that \(\bf tan(0) = \cfrac{\textit{opposite side}}{\textit{adjacent side}}\)

Answer 2

gott it :) i just figured it out as you replied....the answer i got was square root 11/5

Answer 3

hmm \(\bf b = \sqrt{144-100}\implies b = \sqrt{44}\\ \quad \\
tan(0) = \cfrac{\sqrt{44}}{10}\)

Answer 4

ohh I see.. what you meant... root only above.. ok :)

Answer 5

haha yes:)

Answer 6

also if \[\cos0=\frac{ 2\sqrt{10} }{ 7 }\] find the value of sin0.......so the triangle would look like for the missing leg i got \[\sqrt{29}\]

Answer 7

\(\bf b = \sqrt{c^2-a^2}\implies b = \sqrt{7^2-(2\sqrt{10})^2}\implies b = \sqrt{7^2-2^2(\sqrt{10})^2}\\ \quad \\
b = \sqrt{7^2-4\sqrt{10^2}}\implies b = \sqrt{49-4\cdot 10}\)

Answer 8

ok so for sin i got 3/7

Answer 9

yeap