Solve and model 7/10-3/10

1 answer:

7 0

Hello,

$\frac{7}{10} - \frac{3}{10}$ = 2/5

Solution.

<span> $\frac{7}{10} - \frac{3}{10}$ = The denominators are the same so you subtract the numerator. Your answer should be </span> $\frac{4}{10}$ simplified your answer is $\frac{2}{5}$ .

Have A Good Day!

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Use a 30degrees and -60degrees right triangle to find the exact value of the following trigonometric expression. Cot 30 degrees=

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The dirty little secret about trig is students mostly just need to understand the 45/45/90 right isosceles triangle and the 30/60/90 triangle, which is half an isosceles triangle.

I'm sure you have seen it many times already, so I'll remind you in the 30/60/90 right triangle the sides opposite those angles are respectively in the ratio $1: \sqrt{3}:2$ . Remember the smallest side is always opposite the smallest angle, etc. These make a right triangle because

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So in this hypotenuse 2 triangle, the length 1 side is opposite the 30 degree angle and the $\sqrt{3}$ side is adjacent to it.

We could jump right to the answer but let's enumerate all the trig functions. Really you should memorize the sine and cosine so you can get these quickly.

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$\sin 30^\circ = \dfrac{ \textrm{opposite} }{ \textrm{hypotenuse} } = \dfrac{1}{2}$

$\tan 30^\circ = \dfrac{ \textrm{opposite} }{ \textrm{adjacent} }= \dfrac{1}{\sqrt{3}} =\dfrac{\sqrt{3}}{3}$

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$\sec 30^\circ = \dfrac{ \textrm{hypotenuse} }{ \textrm{adjacent}}= \dfrac{2}{\sqrt{3}}$

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We want the cotangent, another way to get it is

$\cot 30^\circ = \dfrac{\cos 30^\circ}{\sin 30^\circ} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}$

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