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Liula [17]
1 year ago
13

7.For the given triangle find the exact value for tan(A). Assume angle C is a right angle.

Mathematics
1 answer:
Vsevolod [243]1 year ago
6 0

tan A = 2/7 (option C)

Explanation:

Angle = A

opposite = side opposite the angle = BC

opposite = 2

adjacent = AC

AC = 7

hypotenuse not given

Since angle C is a right angle, we would apply trigonometry ratio SOHCAHTOA

We are looking for tan A, tan ratio (TOA):

\tan angle=\frac{opposite}{adjacent}\tan \text{ A = }\frac{2}{7}\text{ (}option\text{ C)}

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Find sin(a+b) if tan(a)=7/24 where a is in the third quadrant
ludmilkaskok [199]
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>)
</span>
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 is
sin (a+b)=-36/325

8 0
3 years ago
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Answer:

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Question:<br> 15-4+ (7 - 5)
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Answer:

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Step-by-step explanation:

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