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Select the answer which is equivalent to the given expression using your calculator. tan A=(13)/(84)" and "cos B=(72)/(75) Angles A and B are in Quadrant I. Find sin(A+B). (-828)/(6375) (2700)/(6375) (5775)/(6375) (6321)/(6375)

Select the answer which is equivalent to the given expression using your calculator.\newlinetan⁡A=1384 and cos⁡B=7275 \tan A=\frac{13}{84} \text { and } \cos B=\frac{72}{75} \newlineAngles A and B are in Quadrant I.\newlineFind sin⁡(A+B) \sin (A+B) .\newline−8286375 \frac{-828}{6375} \newline27006375 \frac{2700}{6375} \newline57756375 \frac{5775}{6375} \newline63216375 \frac{6321}{6375}

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Q. Select the answer which is equivalent to the given expression using your calculator.\newlinetan⁡A=1384 and cos⁡B=7275 \tan A=\frac{13}{84} \text { and } \cos B=\frac{72}{75} \newlineAngles A and B are in Quadrant I.\newlineFind sin⁡(A+B) \sin (A+B) .\newline−8286375 \frac{-828}{6375} \newline27006375 \frac{2700}{6375} \newline57756375 \frac{5775}{6375} \newline63216375 \frac{6321}{6375}
  1. Find sine values: Use the given values to find sin⁡A\sin A and sin⁡B\sin B. We are given tan⁡A=1384\tan A = \frac{13}{84} and cos⁡B=7275\cos B = \frac{72}{75}. Since AA and BB are in Quadrant I, both sine values will be positive. To find sin⁡A\sin A, we use the Pythagorean identity: sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. We can find cos⁡A\cos A using the given tan⁡A\tan A value, as sin⁡B\sin B00. Let's find cos⁡A\cos A first: sin⁡B\sin B22. We have the opposite side (sin⁡B\sin B33) and the adjacent side (sin⁡B\sin B44), but we need the hypotenuse for cos⁡A\cos A. Using the Pythagorean theorem, we find the hypotenuse (sin⁡B\sin B66) for triangle AA: sin⁡B\sin B88 sin⁡B\sin B99 tan⁡A=1384\tan A = \frac{13}{84}00 tan⁡A=1384\tan A = \frac{13}{84}11 tan⁡A=1384\tan A = \frac{13}{84}22 tan⁡A=1384\tan A = \frac{13}{84}33 Now, tan⁡A=1384\tan A = \frac{13}{84}44. Using the Pythagorean identity, we can find sin⁡A\sin A: tan⁡A=1384\tan A = \frac{13}{84}66 tan⁡A=1384\tan A = \frac{13}{84}77 tan⁡A=1384\tan A = \frac{13}{84}88 tan⁡A=1384\tan A = \frac{13}{84}99 cos⁡B=7275\cos B = \frac{72}{75}00 cos⁡B=7275\cos B = \frac{72}{75}11 cos⁡B=7275\cos B = \frac{72}{75}22
  2. Find sine B: Find sin⁡B\sin B using the given cos⁡B\cos B. We are given cos⁡B=7275\cos B = \frac{72}{75}. To find sin⁡B\sin B, we use the Pythagorean identity again: sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1 sin⁡2B=1−cos⁡2B\sin^2 B = 1 - \cos^2 B sin⁡2B=1−(7275)2\sin^2 B = 1 - \left(\frac{72}{75}\right)^2 sin⁡2B=1−51845625\sin^2 B = 1 - \frac{5184}{5625} sin⁡2B=5625−51845625\sin^2 B = \frac{5625 - 5184}{5625} sin⁡2B=4415625\sin^2 B = \frac{441}{5625} cos⁡B\cos B00 cos⁡B\cos B11 cos⁡B\cos B22
  3. Use sine addition formula: Use the sine addition formula to find sin⁡(A+B)\sin(A+B). The sine addition formula is sin⁡(A+B)=sin⁡A⋅cos⁡B+cos⁡A⋅sin⁡B\sin(A+B) = \sin A \cdot \cos B + \cos A \cdot \sin B. We already have sin⁡A=1385\sin A = \frac{13}{85} and sin⁡B=725\sin B = \frac{7}{25}. We also have cos⁡A=8485\cos A = \frac{84}{85} and cos⁡B=7275\cos B = \frac{72}{75}. Now we plug these values into the formula: sin⁡(A+B)=(1385)⋅(7275)+(8485)⋅(725)\sin(A+B) = \left(\frac{13}{85}\right) \cdot \left(\frac{72}{75}\right) + \left(\frac{84}{85}\right) \cdot \left(\frac{7}{25}\right) sin⁡(A+B)=9366375+5882125\sin(A+B) = \frac{936}{6375} + \frac{588}{2125} sin⁡(A+B)=9366375+17646375\sin(A+B) = \frac{936}{6375} + \frac{1764}{6375} sin⁡(A+B)=936+17646375\sin(A+B) = \frac{936 + 1764}{6375} sin⁡(A+B)=sin⁡A⋅cos⁡B+cos⁡A⋅sin⁡B\sin(A+B) = \sin A \cdot \cos B + \cos A \cdot \sin B00

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