1 answer:
Answer:
Cos A=5/13
we have
Cos² A=
25/169=1-Sin²A
sin²A=1-25/169
sin²A=144/169
Sin A=
again
Tan B=4/3
P/b=4/3
p=4
b=3
h=
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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