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havianos
ERT
logal systems
lit.
markeling
precale
(8) Sthoplogr

x
4
A
eppss schoology com
PSS Studertst
ertantst
Tangent Sum/Difference Check
3 of 4
POSSIBLE POINTS: 3
Find the value of
tan(a+b) given sia
a=(3)/(15) where
0 < a < (j)/(2) and
cos b=(4)/(5) where
(1)/(2) < b < x. It is suggested that you draw two separate triangles.
Remember your Pythagorean triples!

tan(a+b)=

◻ (Blank 1 is the numerator and
blank 2 is the denominator. Be sure to put in a negative sign i needed but no spaces')

havianos $\newline$ ERT $\newline$ logal systems $\newline$ lit. $\newline$ markeling $\newline$ precale $\newline$ ( $8$ ) Sthoplogr $\newline$ $x$ $\newline$ $4$ $\newline$ A $\newline$ eppss schoology com $\newline$ PSS Studertst $\newline$ ertantst $\newline$ Tangent Sum/Difference Check $\newline$ $3$ of $4$ $\newline$ POSSIBLE POINTS: $3$ $\newline$ Find the value of $\tan (a+b)$ given sia $a=\frac{3}{15}$ where $0<a<\frac{j}{2}$ and $\cos b=\frac{4}{5}$ where $\frac{1}{2}<b<x$ . It is suggested that you draw two separate triangles. $\newline$ Remember your Pythagorean triples! $\newline$ $\tan (a+b)=$ $\newline$ $\square$ (Blank $1$ is the numerator and $\newline$ blank $2$ is the denominator. Be sure to put in a negative sign i needed but no spaces')

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Calculus

Intermediate Value Theorem

Full solution

Q. havianos

\newline

ERT

\newline

logal systems

\newline

lit.

\newline

markeling

\newline

precale

\newline

(

8

) Sthoplogr

\newline

x

\newline

4

\newline

\newline

eppss schoology com

\newline

PSS Studertst

\newline

ertantst

\newline

Tangent Sum/Difference Check

\newline

3

4

\newline

POSSIBLE POINTS:

3

\newline

Find the value of

\tan (a+b)

given sia

a=\frac{3}{15}

where

0<a<\frac{j}{2}

and

\cos b=\frac{4}{5}

where

\frac{1}{2}<b<x

. It is suggested that you draw two separate triangles.

\newline

Remember your Pythagorean triples!

\newline

\tan (a+b)=

\newline

\square

(Blank

1

is the numerator and

\newline

blank

2

is the denominator. Be sure to put in a negative sign i needed but no spaces')

Calculate $\sin(a)$ : Calculate $\sin(a)$ and reduce the fraction. $\newline$ $\sin(a) = \frac{3}{15} = \frac{1}{5}$ .
Find missing side: Find the missing side of the right triangle for angle $a$ using the Pythagorean theorem. $\sin(a) = \frac{opposite}{hypotenuse}$ , so the opposite side is $1$ and the hypotenuse is $5$ . The adjacent side is $\sqrt{hypotenuse^2 - opposite^2} = \sqrt{5^2 - 1^2} = \sqrt{25 - 1} = \sqrt{24}$ .
Simplify $\sqrt{24}$ : Simplify $\sqrt{24}$ to get the length of the adjacent side. $\newline$ $\sqrt{24} = 2\sqrt{6}$ .
Calculate $\cos(a)$ : Calculate $\cos(a)$ using the adjacent side and hypotenuse. $\cos(a) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2\sqrt{6}}{5}$ .
Find missing side: Find the missing side of the right triangle for angle $b$ using the Pythagorean theorem. $\cos(b) = \frac{\text{adjacent}}{\text{hypotenuse}}$ , so the adjacent side is $4$ and the hypotenuse is $5$ . The opposite side is $\sqrt{\text{hypotenuse}^2 - \text{adjacent}^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9}$ .
Simplify $\sqrt{9}$ : Simplify $\sqrt{9}$ to get the length of the opposite side. $\newline$ $\sqrt{9} = 3$ .
Calculate $\sin(b)$ : Calculate $\sin(b)$ using the opposite side and hypotenuse. $\sin(b) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$ .
Use $\tan(a+b)$ formula: Use the formula for $\tan(a+b) = \frac{\sin(a)\cos(b) + \cos(a)\sin(b)}{\cos(a)\cos(b) - \sin(a)\sin(b)}$ . Substitute the values found for $\sin(a)$ , $\cos(a)$ , $\sin(b)$ , and $\cos(b)$ . $\tan(a+b) = \frac{(\frac{1}{5})(\frac{4}{5}) + (\frac{2\sqrt{6}}{5})(\frac{3}{5})}{(\frac{2\sqrt{6}}{5})(\frac{4}{5}) - (\frac{1}{5})(\frac{3}{5})}$ .
Simplify numerator and denominator: Simplify the numerator and denominator. $\tan(a+b) = \frac{(\frac{4}{25} + \frac{6\sqrt{6}}{25})}{(\frac{8\sqrt{6}}{25} - \frac{3}{25})}$ .
Correct denominator: Combine terms in the numerator and denominator. $\newline$ $\tan(a+b) = \frac{4 + 6\sqrt{6}}{8\sqrt{6} - 3}$ .
Correct denominator: Combine terms in the numerator and denominator. $\newline$ $\tan(a+b) = \frac{4 + 6\sqrt{6}}{8\sqrt{6} - 3}$ .Realize there's a mistake in the previous step; the denominator should be simplified correctly. $\newline$ Correct the denominator: $\frac{8\sqrt{6}}{25} - \frac{3}{25} = \frac{8\sqrt{6} - 3}{25}$ .