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2 years ago

If sin x = 12÷13 where 0° <x<90°find the value of 1-cos ×cos( x)

Mathematics

1 answer:

s2008m [1.1K]2 years ago

4 0

Answer:

This is how tackle this type of problem:

Draw a right triangle with the indicated sides: side opposite the angle = 12 and hypotenuse = 13

Then, calculate the 3rd side by Pythagoras: x2 + 122 = 132

x2 = 169 - 144

x2 = 25

x = 5

Now, we have all 3 sides of our diagram triangle, so we can find all the trig ratios for the angle.

Next, we use the double angle formula to simplify cos(x + 30) = cos x cos 30 - sin x sin 30

= 5/13 . √3/2 - 12/13 . 1/2

= 5√3 /26 - 6/13

= (5√3 - 12) / 26

Step-by-step explanation:

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7+4(r+3)=9r-6 show work pls AND HELP ASAP

ivanzaharov [21]

$\: \: \: \: \: \:$

$= \: \: \: \: r = 5$

Step-by-step explanation:

$7 + 4(r + 3) = 9r - 6$

distribute 4 through the parentheses

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add numbers

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move the variable to the left hand side and change its sign

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move the constant to the right hand side and change its sign

$4r - 9 = - 6 - 19$

collect like terms

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divide both sides of the equation by -5

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<h2>hope it helps</h2>

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2 years ago

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Answer:

Step-by-step explanation:

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2 years ago

If sin x = 12÷13 where 0° <x<90°find the value of 1-cos ×cos( x)​

If sin x = 12÷13 where 0° <x<90°find the value of 1-cos ×cos( x)