1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bixtya [17]
3 years ago
14

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

os B = 12/13
-220/221
-140/221
140/221
220/221

If A and B are two angles in standard position in Quadrant I, find cos( A -B ) for the given function values. sin A = 3/5 and cos B= 12/37
153/185
57/185
-57/185
-153/185

If A and B are two angles in standard position in Quadrant I, find cos( A - B) for the given function values.
sin A = 15/17 and cos = 3/5
-84/85
-36/85
36/85
84/85

If A and B are two angles in standard position in Quadrant I, find cos( A + B) for the given function values.
sin A = 15/17 and cos = 3/5
-220/221
-140/221
140/221
220/221

If A and B are two angles in standard position in Quadrant I, find cos( A - B) for the given function values.
sin A = 4/5 and cos = 5/13
-33/65
33/65
-63/65
63/65

If A and B are two angles in standard position in Quadrant I, find cos( A + B) for the given function values.
sin A = 3/5 and cos = 12/37
153/185
57/185
-57/185
-153/185
Mathematics
1 answer:
horsena [70]3 years ago
4 0

Answer:

Part 1) cos(A + B) = \frac{140}{221}

Part 2) cos(A - B) = \frac{153}{185}

Part 3) cos(A - B) = \frac{84}{85}

Part 4) cos(A + B) = -\frac{36}{85}

Part 5) cos(A - B) = \frac{63}{65}

Part 6) cos(A+ B) = -\frac{57}{185}

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

sin(A)=\frac{8}{17}

cos(B)=\frac{12}{13}

Determine cos (A+B)

we know that

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{8}{17})^2=1

cos^2(A)+\frac{64}{289}=1

cos^2(A)=1-\frac{64}{289}

cos^2(A)=\frac{225}{289}

cos(A)=\pm\frac{15}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{15}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{13})^2=1

sin^2(B)+\frac{144}{169}=1

sin^2(B)=1-\frac{144}{169}

sin^2(B)=\frac{25}{169}

sin(B)=\pm\frac{25}{169}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{5}{13}

step 3

Find cos(A+B)

substitute in the formula

cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}

cos(A + B) = \frac{180}{221}-\frac{40}{221}

cos(A + B) = \frac{140}{221}

Part 2) we have

sin(A)=\frac{3}{5}

cos(B)=\frac{12}{37}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{3}{5})^2=1

cos^2(A)+\frac{9}{25}=1

cos^2(A)=1-\frac{9}{25}

cos^2(A)=\frac{16}{25}

cos(A)=\pm\frac{4}{5}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{4}{5}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{37})^2=1

sin^2(B)+\frac{144}{1,369}=1

sin^2(B)=1-\frac{144}{1,369}

sin^2(B)=\frac{1,225}{1,369}

sin(B)=\pm\frac{35}{37}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{35}{37}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}

cos(A - B) = \frac{48}{185}+\frac{105}{185}

cos(A - B) = \frac{153}{185}

Part 3) we have

sin(A)=\frac{15}{17}

cos(B)=\frac{3}{5}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}

cos(A - B) = \frac{24}{85}+\frac{60}{85}

cos(A - B) = \frac{84}{85}

Part 4) we have

sin(A)=\frac{15}{17}        

cos(B)=\frac{3}{5}

Determine cos (A+B)

we know that    

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}      

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A+B)

substitute in the formula    

cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}

cos(A + B) = \frac{24}{85}-\frac{60}{85}

cos(A + B) = -\frac{36}{85}

Download odt
You might be interested in
PLEASE HELP ASAP WILL GIVE BRAINLEIST!!!
lyudmila [28]

Answer:

$ 3125.00

Step-by-step explanation:

P=\frac{SI * 100 }{R*T}=

\frac{500*100 }{4*4} = \frac{50000}{16} = $3125.00

7 0
3 years ago
Derek makes $8ban hour. Last week his paycheck was $280, before taxes. How many hours did he work last week?
Ber [7]

Answer:

35 hours that week

Step-by-step explanation:

280÷8=35

35 hours that week

3 0
3 years ago
Read 2 more answers
What’s a line equation passing through (14,-8) and (24,-3)
Nataly_w [17]

\bf (\stackrel{x_1}{14}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{24}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-8)}}}{\underset{run} {\underset{x_2}{24}-\underset{x_1}{14}}}\implies \cfrac{-3+8}{10}\implies \cfrac{5}{10}\implies \cfrac{1}{2}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-8)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{14}) \\\\\\ y+8=\cfrac{1}{2}x-7\implies y=\cfrac{1}{2}x-15

4 0
3 years ago
Which of the following equations is represented by the graph?
Greeley [361]

Answer:

••••••••••••••••••••The answer is Y=2/3x

6 0
2 years ago
Read 2 more answers
Could someone pls explain this step by step? Really simplify it pls because I don't understand it at all.
Lunna [17]

Answer:

  • y = 2x -6
  • y = -x +2
  • y = 4/5x -2
  • y = -3/2x +4

Step-by-step explanation:

The slope-intercept form of the equation for a line is ...

  y = mx +b . . . . . where m is the slope, and b is the y-intercept

For these equations, writing them in slope-intercept form means "solve for y." You always do any "solve for ..." problem by undoing the operations that are done to the variable. Addition is undone by adding the opposite. Multiplication is undone by division.

As always, whatever operation we perform on one side of the equal sign must also be performed on the other side. That is how the equal sign remains valid.

__

<h3>a)</h3>

  1/2y = x -3

All we need to do is get rid of the coefficient of y. We can do that by dividing by 1/2, or multiplying by 2. The latter is easier to see, perhaps.

 2(1/2y) = 2(x -3)

  y = 2x +2(-3)

  y = 2x -6

__

<h3>b)</h3>

  x +y = 2

Here, we need to undo the addition of x. We do that by adding (-x) to both sides of the equation.

  -x +x +y = -x +2

  y = -x +2

__

<h3>c)</h3>

  4x -5y -10 = 0

This is a combination of the above operations. We can separate the y-term from the others by either subtracting the others, or adding the opposite of the y-term. We choose the latter, because that will give a positive coefficient for y. Adding 5y to both sides gives ...

  4x -5y +5y -10 = 5y

  5y = 4x -10 . . . . . . . . simplify and put y on the left

Now, we need to divide by the coefficient of y. We do that to both sides of the equation.

  5y/5 = (4x -10)/5

  y = 4/5x -10/5

  y = 4/5x -2

__

<h3>d)</h3>

  3x = -2y +8

We have a constant with the y-term. To get rid of it, we add its opposite to both sides.

  3x -8 = -2y +8 -8

  3x -8 = -2y

As before, we divide by the coefficient of y:

  (3x -8)/(-2) = (-2y)/(-2)

  -3/2x +4 = y

  y = -3/2x +4

3 0
1 year ago
Other questions:
  • Barry is training to be a gymnast. He increases the number of push-ups he does each week by following a number pattern. The numb
    5·2 answers
  • Andrew solves 6x2=12x by using the quadratic formula. What values does he use for a, b, and c? a = (Put the number only in the b
    9·1 answer
  • Which of the following represent(s) an equation of the line passing through the points A(5, 6) and B(4, 8). Select all that appl
    12·1 answer
  • What is the area of a triangle with base of 20 feet and a vertical height
    13·1 answer
  • PLS HELP I WILL GUVE POINT
    6·2 answers
  • *Just use 10° for your angle *
    15·1 answer
  • Can 4,5,6 form any triangle
    6·1 answer
  • Write a rule relating to the minutes, x, to the number of pages, y. Then find the number of pages when the minutes equal 72.
    8·1 answer
  • Pets Plus and Pet Planet are having a sale on the same aquarium. At Pets Plus the aquarium is on sale for 30% off the original p
    14·1 answer
  • Determine if each set of side lengths could be used to create a triangle. *Explain your reasoning.*
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!