The complete question in the attached figure
we have that
tan a=7/24 a----> III quadrant
cos b=-12/13 b----> II quadrant
sin (a+b)=?
we know that
sin(a + b) = sin(a)cos(b) + cos(a)sin(b<span>)
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step 1
find sin b
sin²b+cos²b=1------> sin²b=1-cos²b----> 1-(144/169)---> 25/169
sin b=5/13------> is positive because b belong to the II quadrant
step 2
Find sin a and cos a
tan a=7/24
tan a=sin a /cos a-------> sin a=tan a*cos a-----> sin a=(7/24)*cos a
sin a=(7/24)*cos a------> sin²a=(49/576)*cos²a-----> equation 1
sin²a=1-cos²a------> equation 2
equals 1 and 2
(49/576)*cos²a=1-cos²a---> cos²a*[1+(49/576)]=1----> cos²a*[625/576]=1
cos²a=576/625------> cos a=-24/25----> is negative because a belong to III quadrant
cos a=-24/25
sin²a=1-cos²a-----> 1-(576/625)----> sin²a=49/625
sin a=-7/25-----> is negative because a belong to III quadrant
step 3
find sin (a+b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin a=-7/25
cos a=-24/25
sin b=5/13
cos b=-12/13
so
sin (a+b)=[-7/25]*[-12/13]+[-24/25]*[5/13]----> [84/325]+[-120/325]
sin (a+b)=-36/325
the answer issin (a+b)=-36/325
I think this is it not sure tho
Answer:
In the given figure, line segment is constructed on the triangle parallel to line segment . ... This line can then be considered as a transversal of these parallel lines. Angle and angle are therefore corresponding angles. And as they're corresponding, this means that they're equal.
Answer:
13
Step-by-step explanation:
15 - 4 + 7 - 5 = 13
Simply expand the brackets and subtract and add accordingly.
• Use slope to graph linear equations in two variables.
• Find the slope of a line given two points on the line.
• Write linear equations in two variables.
• Use slope to identify parallel and perpendicular lines.
• Use slope and linear equations in two variables to model and solve real-life problems.
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