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Ray Of Light [21]
2 years ago
13

Please help me with this.

Mathematics
1 answer:
zheka24 [161]2 years ago
7 0

The sin A is equal to 12/13 and the tan (A) is equal to 12/5.

<h3>RIGHT TRIANGLE</h3>

A triangle is classified as a right triangle when it presents one of your angles equal to 90º.  The greatest side of a right triangle is called hypotenuse. And, the other two sides are called cathetus or legs.

The math tools applied for finding angles or sides in a right triangle are the trigonometric ratios or the Pythagorean Theorem.

The Pythagorean Theorem says: hypotenuse^2=(leg_1)^2+(leg_2)^2. And the main trigonometric ratios are:

sin(\beta )= \frac{opposite\;leg}{hypotenuse} \\ \\ cos(\beta )= \frac{adjacent\;leg}{hypotenuse} \\ \\ tan(\beta )= \frac{opposite\;leg}{adjacent\;leg} \\ \\

The question gives cos (A)=5/13. If cos (A) is represented by the quotient between the adjacent leg and the hypotenuse, you have:

adjacent leg=5

hypotenuse=13

Therefore, you can find the opposite leg of A from Pythagorean Theorem, see below.

hypotenuse^2=(leg_1)^2+(leg_2)^2\\ \\ 13^2=5^2+(leg_2)^2\\ \\ 169=25+(leg_2)^2\\ \\ 144=(leg_2)^2\\ \\ leg_2=12

Thus, the opposite leg is equal to 12. Now, you can find sin (A) since:

sin(A)= \frac{opposite\;leg}{hypotenuse}=\frac{12}{13}

Finally, you can find the tan (A) since:

tan(a )= \frac{opposite\;leg}{adjacent\;leg}=\frac{12}{5}

Learn more about trigonometric ratios here:

brainly.com/question/11967894

#SPJ1

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ANSWER

Find out the how many 3, 4, and 5 point questions could there be.

To proof

Let us assume that the 3 points based question be = x

Let us assume that the 4 points based question be = y

Let us assume that the 5 points based question be = z

As given

An instructor wants to write a test with 25 questions

than the equation become in the form

x + y + z = 25

he quiz to be worth a total of 100 points.

than the equation is becomes

3x + 4y + 5z =100

As given

He wants the number of 3-point questions to be 4 less than the number of 4-point questions

x = y -4

Than the three equation are

x + y + z = 25 ,3x + 4y + 5z =100 and x = y -4

put  x = y -4 in the x + y + z = 25 ,3x + 4y + 5z =100

than

2y + z = 29, 7y +5z= 112

multiply 2y + z = 29 by 5 and subtracted 7y +5z= 112

10 y -7y + 5y -5y = 145 -112

3y = 33

y = \frac{33}{3}

y =11

put in the  x = y -4

x = 11-4

x= 7

put the value of x ,y in the x + y + z = 25

7 + 11 +z =25

18 +z = 25

z = 25 -18

z=7

therefore

numbers of the 3 point question be = 7

numbers of the 4 point question be = 11

numbers of the 5 point question be = 7

Hence proved



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