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yanalaym [24]
3 years ago
11

E

Mathematics
1 answer:
Allushta [10]3 years ago
8 0

Answer:

Step-by-step explanation:

We can use Hypotenuse Theorem for this:

For triangle ABC,

AC²=CB²+BA²

9²=CB²+3²

Switch the sides

-CB²=3² -9²

-CB²= 9-81

-CB²= -72

the signs cut each other

CB²=72

the square changes sides

CB=√72

CB= 8.49 cm

Now triangle CBD

We have the length of CB but not the other two whereas we have the angle

It will be sin

As we know sin= O/H

Here O is CB whereas H is CD

Put the values

sin 50= 8.49/H

H= 8.49 ÷sin 50

Calculate

H=11.1 cm

As we know H is CD so CD is 11.1 cm

Hope it helps

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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

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Find the angle between the given vectors to the nearest tenth of a degree. u = &lt;-5, -4&gt;, v = &lt;-4, -3&gt; (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

u.v =|u|(|v|)cosx , where x is angle between u & v !

⇒ u.v =|u|(|v|)cosx

⇒ cosx =\frac{u.v}{|u|(|v|)}

Now , u.v = (-5i-4j)(-4i-3j)

⇒ u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)

⇒ u.v = 20+12            { i(j) = j(i) =0  }

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Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

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⇒ cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})

⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
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