Question 1209566
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Please help. Thank you in advance.

Given tan 𝛼 =7/24, 𝛼 in the third quadrant, sin 𝛽 =2/√13, 𝛽 in the second quadrant, find:
a) the quadrant containing 𝛼 + 𝛽
b) the quadrant containing 𝛼 - 𝛽
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<pre>
Notice that 7^2 + 24^ = 625 = 25^2.


So, you may think that 

    tan(a) = {{{7/24}}} in the third quadrant is  {{{((-7/25))/((-24/25))}}} = {{{sin(a)/cos(a)}}}.


In other words, you may think that sin(a) = {{{-7/25}}},  cos(a) = {{{-24/25}}}.

    The signs at  sin(a)  and  cos(a)  do agree that  "a"  is in the third quadrant.



Next, you are given that sin(b) = {{{2/sqrt(13)}}} in the second quadrant; so, you can calculate 

    cos(b) = {{{-sqrt(1-sin^2(b))}}} = {{{-sqrt(1-4/13)}}} = {{{-sqrt((13-4)/13)}}} = {{{-sqrt(9/13)}}} = {{{-3/sqrt(13)}}}.

    The sign at  cos(b)  does agree that  "b"  is in the second quadrant.



Now, as you know  sin(a) {{{-7/25}}},  cos(a) = {{{-24/25}}}, sin(b) = {{{2/sqrt(13)}}},  cos(b) = {{{-3/sqrt(13)}}},  you can calculate


    sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b) = {{{(-7/25)*(-3/sqrt(13))}}} + {{{(-24/25)*(2/sqrt(13))}}} = {{{21/(25*sqrt(13))}}} - {{{48/(25*sqrt(13))}}} = 

             = {{{(21-48)/(25*sqrt(13))}}} = {{{-27/(25*sqrt(13))}}},


    cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) = {{{(-24/25)*(-3/sqrt(13))}}} - {{{(-7/25)*(2/sqrt(13))}}} = {{{72/(25*sqrt(13))}}} - {{{(-14/(25*sqrt(13)))}}} = 

             = {{{(72+14)/(25*sqrt(13))}}} = {{{86/(25*sqrt(13))}}}.


Thus, sin(a+b) is a negative real number;  cos(a+b) is a positive real number.


It means that angle a+b is in fourth quadrant.


<U>ANSWER</U>.  Angle a+b is in fourth quadrant.
</pre>

Part &nbsp;(a) &nbsp;is solved completely.


For part &nbsp;(b), &nbsp;calculate &nbsp;sin(a-b) &nbsp;and &nbsp;cos(a-b)  similarly; &nbsp;then make a conclusion about angle &nbsp;a-b.

You just have a &nbsp;TEMPLATE &nbsp;for it.