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Reptile [31]
2 years ago
13

Find sin(a+B)

Mathematics
1 answer:
diamong [38]2 years ago
6 0

Answer:

Cos A=5/13

we have

Cos² A=1-Sin ²A

25/169=1-Sin²A

sin²A=1-25/169

sin²A=144/169

Sin A=\sqrt{144/169}=12/13

again

Tan B=4/3

P/b=4/3

p=4

b=3

h=\sqrt{3²+4²}=5

Now

Sin B=p/h=4/5

in IV quadrant sin angle is negative so

Sin B=-4/5

CosB=b/h=3/5

Now

<u>S</u><u>i</u><u>n</u><u>(</u><u>A</u><u>+</u><u>B</u><u>)</u><u>:</u><u>s</u><u>i</u><u>n</u><u>A</u><u>c</u><u>o</u><u>s</u><u>B</u><u>+</u><u>C</u><u>o</u><u>s</u><u>A</u><u>s</u><u>i</u><u>n</u><u>B</u>

<u>n</u><u>o</u><u>w</u><u> </u>

<u>substitute</u><u> </u><u>value</u>

<u>Sin(A+B):</u>12/13*3/5+5/13*(-4/5)=36/65-4/13

<u>=</u><u>1</u><u>6</u><u>/</u><u>6</u><u>5</u><u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u>

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