Answer:
Cos A=5/13
we have
Cos² A=![1-Sin ²A](https://tex.z-dn.net/?f=%201-Sin%20%C2%B2A)
25/169=1-Sin²A
sin²A=1-25/169
sin²A=144/169
Sin A=![\sqrt{144/169}=12/13](https://tex.z-dn.net/?f=%20%5Csqrt%7B144%2F169%7D%3D12%2F13)
again
Tan B=4/3
P/b=4/3
p=4
b=3
h=![\sqrt{3²+4²}=5](https://tex.z-dn.net/?f=%5Csqrt%7B3%C2%B2%2B4%C2%B2%7D%3D5)
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>
Y-y1=m(x-x1)
y+5=14/3* (x+3)
y+5= (14/3)x+14
y=(14/3)x+9
(14/3)x+9-y=0 /*3
14x-3y+27=0
Answer:
2
Step-by-step explanation:
/2= 8
8+6= 14
4+3=7
14/7= 2