I NEED HELP ASAP jjjuyyy

Mathematics

1 answer:

MissTica2 years ago

3 0

Yeah it’s defiantly then first but maybe

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I have to use trigonometric identities to solve. But I’m having trouble finding the values of cos A and sin B. Can anyone help m

katrin [286]

let's notice something, angles α and β are both in the I Quadrant, and on the first quadrant the x-coordinate/cosine and y-coordinate/sine are both positive.

$\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\alpha)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{17^2-15^2}=a\implies \pm\sqrt{64}=a\implies \pm 8 = a\implies \stackrel{I~Quadrant}{\boxed{+8=a}} \\\\[-0.35em] ~\dotfill\\\\ cos(\beta)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{\textit{I~Quadrant}}{\boxed{+4=b}} \\\\[-0.35em] ~\dotfill$

$\bf cos(\alpha - \beta)=\stackrel{cos(\alpha)}{\left( \cfrac{8}{17} \right)}\stackrel{cos(\beta)}{\left( \cfrac{3}{5} \right)}+\stackrel{sin(\alpha)}{\left( \cfrac{15}{17} \right)}\stackrel{sin(\beta)}{\left( \cfrac{4}{5} \right)}\implies cos(\alpha - \beta)=\cfrac{24}{85}+\cfrac{60}{85} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos(\alpha - \beta)=\cfrac{84}{85}~\hfill$

5 0

3 years ago

given triangle GHI with G(4,-3), H(-4,2) and I(2,4), find the perpendicular bisector of HI in standard form.

nekit [7.7K]

The perpendicular bisector would be the line that connects with the line HI at a 90 degree angle and in the mid-point. This means that we would have to find the slope of the line HI and take the opposite. The slope of HI is 1/3. This means that the slope of the perpendicular line is -3. Now we must solve for the line that crosses through the midpoint. y=mx+b we know that m=-3 and that the midpoint is at -1,3 so x=-1 and y=3. 3=(-3)-1+b. We then solve for b and find it to be 0. So the final answer is y= -3x

8 0

2 years ago

Read 2 more answers

Which equation represents the line that passes

viktelen [127]

Answer:

1) y=3x+1

Step-by-step explanation:

passes through the points

(-1.-2) and (3, 10)

7 0

2 years ago