Mathematics, 17.02.2020 00:47, herbal420medici

Find the EXACT value of cos(A+B) if sinA=−8/17 where A is in Quadrant III and sec B =−5/4 where B is in Quadrant II. Assume all angles are measured from standard position. cos(A+B) =

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answered: guzal5733

$\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)$

we know that angle A is in the III Quadrant, where sine is negative and cosine is negative.

we also know that angle B is in the II Quadrant, where sine is positive and cosine is negative.

the hypotenuse is never negative, since it's simply a radius length, so in a fraction, the hypotenuse will be the positive value.

$\bf sin(A)=\cfrac{\stackrel{opposite}{-8}}{\underset{hypotenuse}{17}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{17^2-(-8)^2}=a\implies \pm\sqrt{225}=a\implies \pm 15 = a\implies \stackrel{III~Quadrant}{-15=a} \\\\\\ therefore\qquad \qquad cos(A)=\cfrac{-15}{17} \\\\[-0.35em] ~\dotfill$

$\bf sec(B)=\cfrac{\stackrel{hypotenuse}{5}}{\underset{adjacent}{-4}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{5^2-(-4)^2}=b\implies \pm\sqrt{9}=b \\\\\\ \pm 3 = b\implies \stackrel{II~Quadrant}{3=b}~\hfill cos(B)=\cfrac{\stackrel{adjacent}{-4}}{\underset{hypotenuse}{5}}~\hfill sin(B)=\cfrac{\stackrel{opposite}{3}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill$

$\bf cos(A+B)=cos(A)cos(B)-sin(A)sin(B) \\\\\\ cos(A+B)=\cfrac{-15}{17}\cdot \cfrac{-4}{5}-\left(\cfrac{-8}{17}\cdot \cfrac{3}{5} \right)\implies cos(A+B)=\cfrac{60}{85}-\left( \cfrac{-24}{85} \right) \\\\\\ cos(A+B)=\cfrac{60}{85}+\cfrac{24}{85}\implies cos(A+B)=\cfrac{84}{85}$

answered: Guest

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

answered: Guest

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2 polynomial long division

dividing : x4+5x3-4x2-8x+6

("dividend")

by : x-1 ("divisor")

dividend x4 + 5x3 - 4x2 - 8x + 6

- divisor * x3 x4 - x3

remainder 6x3 - 4x2 - 8x + 6

- divisor * 6x2 6x3 - 6x2

remainder 2x2 - 8x + 6

- divisor * 2x1 2x2 - 2x

remainder - 6x + 6

- divisor * -6x0 - 6x + 6

remainder 0

quotient : x3+6x2+2x-6 remainder: 0

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Find the EXACT value of cos(A+B) if sinA=−8/17 where A is in Quadrant III and sec B =−5/4 where B is...