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Mathematics : asked on dreparksw
 18.01.2021

Which trigonometric function is represented by the ratio opposite leg hypotenuse ?

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15.12.2022, solved by verified expert
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Bhaskar Singh Bora
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Answer: Sin

We will call the ratio of the opposite side of a right triangle to the hypotenuse the sine and give it the symbol sin. The ratio of the adjacent side of a right triangle to the hypotenuse is called the cosine and given the symbol cos.

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Faq

Mathematics
In δrst, which trigonometric function is represented by the ratio st rs ?
Step-by-step answer
P Answered by Master

the inverse of Hypotenus/adjacent

Step-by-step explanation:

using the trigonometric identity

SOHCAHTOA

where sin = Opposite/ hypotenus

cos = Adjacent / Hypotenus

Tan = Opposite/ Adjacent

in this question we are assuming that ST is the hypotenus

therefore SR  or RT could be the opposite or adjacent depending on where the angle given is placed

the ratio of ST : RS  =  hypotenus / adjacent or hypotenus / opposite

Mathematics
In δrst, which trigonometric function is represented by the ratio st rs ?
Step-by-step answer
P Answered by Specialist

the inverse of Hypotenus/adjacent

Step-by-step explanation:

using the trigonometric identity

SOHCAHTOA

where sin = Opposite/ hypotenus

cos = Adjacent / Hypotenus

Tan = Opposite/ Adjacent

in this question we are assuming that ST is the hypotenus

therefore SR  or RT could be the opposite or adjacent depending on where the angle given is placed

the ratio of ST : RS  =  hypotenus / adjacent or hypotenus / opposite

Mathematics
Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. Rationalize the denominator if applicable. Find tan B when a
Step-by-step answer
P Answered by PhD
Unknown side = 28tan B = 7/24

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. Rationalize the denominator if applicable. Find tan B when a = 96 and c = 100.

Pythagoras theorem states that the square of the hypotenuse side of a right angled triangle is equal to the sum of the square of its other two sides. Mathematically c² = a²+b² where c is the hypotenuse and a,b are the other two sides.

From the question, we are given a = 96 and c = 100, to get the unknown side 'b', we will substitute the given values into the formula above;

c² = a²+b²

100²  = 96² +b²

b²  = 100²  - 96²

b²  = 10,000 - 9216

b²  = 784

b = √784

b = 28

Hence, the unknown length is 28.

To get tanB, we will use the SOH, CAH, TOA trigonometry identity

According to TOA, tan B = opposite/adjacent

tan B = b/a (note that side b is the opposite in this case since the angle we are considering is B)

Given b = 28 and a = 96

tan B = 28/96

tan B = 4*7/4*24

tan B = 7/24

Mathematics
Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. Rationalize the denominator if applicable. Find tan B when a
Step-by-step answer
P Answered by PhD
Unknown side = 28tan B = 7/24

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. Rationalize the denominator if applicable. Find tan B when a = 96 and c = 100.

Pythagoras theorem states that the square of the hypotenuse side of a right angled triangle is equal to the sum of the square of its other two sides. Mathematically c² = a²+b² where c is the hypotenuse and a,b are the other two sides.

From the question, we are given a = 96 and c = 100, to get the unknown side 'b', we will substitute the given values into the formula above;

c² = a²+b²

100²  = 96² +b²

b²  = 100²  - 96²

b²  = 10,000 - 9216

b²  = 784

b = √784

b = 28

Hence, the unknown length is 28.

To get tanB, we will use the SOH, CAH, TOA trigonometry identity

According to TOA, tan B = opposite/adjacent

tan B = b/a (note that side b is the opposite in this case since the angle we are considering is B)

Given b = 28 and a = 96

tan B = 28/96

tan B = 4*7/4*24

tan B = 7/24

Mathematics
Find exact values of the six trigonometric functions for theangle 300 degrees (rationalizing the denominator).
Step-by-step answer
P Answered by Specialist

Since, the six main trigonometric functions are:

Sine (sin)Cosine (cos)Tangent (tan)Secant (sec)Cosecant (csc)Cotangent (cot)

If we have the exact value of any one trigonometric function for a degree then we can find the other function for the same degree as follow.

sin 300° = sin (-60 + 360)° = sin (-60) = - sin 60 = -\frac{\sqrt{3}}{2}

cos 300° = cos (-60 + 360)° = cos (-60) = cos 60 = \frac{1}{2}

tan 300° = \frac{\sin 300^{\circ}}{\cos 300^{\circ}}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}

cot  300° = \frac{1}{\tan 300^{\circ}}=\frac{1}{-\sqrt{3}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}

sec 300° = \frac{1}{\cos 300^{\circ}}=\frac{1}{\frac{1}{2}}=2

csc 300° = \frac{1}{\sin 300^{\circ}}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}

Note : sin (-x) = -sin x and cos (-x) = cos x

Mathematics
Find exact values of the six trigonometric functions for theangle 300 degrees (rationalizing the denominator).
Step-by-step answer
P Answered by Master

Since, the six main trigonometric functions are:

Sine (sin)Cosine (cos)Tangent (tan)Secant (sec)Cosecant (csc)Cotangent (cot)

If we have the exact value of any one trigonometric function for a degree then we can find the other function for the same degree as follow.

sin 300° = sin (-60 + 360)° = sin (-60) = - sin 60 = -\frac{\sqrt{3}}{2}

cos 300° = cos (-60 + 360)° = cos (-60) = cos 60 = \frac{1}{2}

tan 300° = \frac{\sin 300^{\circ}}{\cos 300^{\circ}}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}

cot  300° = \frac{1}{\tan 300^{\circ}}=\frac{1}{-\sqrt{3}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}

sec 300° = \frac{1}{\cos 300^{\circ}}=\frac{1}{\frac{1}{2}}=2

csc 300° = \frac{1}{\sin 300^{\circ}}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}

Note : sin (-x) = -sin x and cos (-x) = cos x

Mathematics
6.which trigonometric function is equivalent to f(x)=sin x? a) f(x)=cos(-x+(3pi/2)) b) f(x)=cos(x+(pi/2)) c) f(x)=cos(-x+(pi/2)) d) f(x)=cos(-x+pi) 7.which trigonometric function requires a domain restriction of [(-pi//2)] to make it invertible? a) f(x)=sin x b) f(x)=cos x c) f(x)=tan x d) f(x)=csc x e) f(x)=sec x f) f(x)=cot x 8.what is the horizontal shift of the function y= 3sin(4x-pi)? a)pi units to the right b)pi/2 units to the right c)pi/3 units to the left d)pi/4 units to the right 9.which is the exact value of cos (-75 degrees)? a) (sqrt6
Step-by-step answer
P Answered by Master

ANSWER

See attachment for questions 6 to 25

y= - 11x - 16

QUESTION 26

The given function is:

f(x) = 4 {x}^{2} + 5x

The slope function is obtained by taking the first derivative of the function with respect to x.

f'(x) = 8x + 5

At (-2,6), x= -2

We plug in the x-value to find the slope of the function.

f'( - 2) = 8( - 2)+ 5

f'( - 2) = - 16+ 5

f'( - 2) = - 11

The equation is given by:

y-y_1=m(x-x_1)

We substitute the point and slope into the formula to get;

y-6= - 11(x- - 2)

y= - 11x - 22 + 6

y= - 11x - 16

The correct answer is D


6.which trigonometric function is equivalent to f(x)=sin x? a) f(x)=cos(-x+(3pi/2)) b) f(x)=cos(x+(
Mathematics
6.which trigonometric function is equivalent to f(x)=sin x? a) f(x)=cos(-x+(3pi/2)) b) f(x)=cos(x+(pi/2)) c) f(x)=cos(-x+(pi/2)) d) f(x)=cos(-x+pi) 7.which trigonometric function requires a domain restriction of [(-pi//2)] to make it invertible? a) f(x)=sin x b) f(x)=cos x c) f(x)=tan x d) f(x)=csc x e) f(x)=sec x f) f(x)=cot x 8.what is the horizontal shift of the function y= 3sin(4x-pi)? a)pi units to the right b)pi/2 units to the right c)pi/3 units to the left d)pi/4 units to the right 9.which is the exact value of cos (-75 degrees)? a) (sqrt6
Step-by-step answer
P Answered by Specialist

ANSWER

See attachment for questions 6 to 25

y= - 11x - 16

QUESTION 26

The given function is:

f(x) = 4 {x}^{2} + 5x

The slope function is obtained by taking the first derivative of the function with respect to x.

f'(x) = 8x + 5

At (-2,6), x= -2

We plug in the x-value to find the slope of the function.

f'( - 2) = 8( - 2)+ 5

f'( - 2) = - 16+ 5

f'( - 2) = - 11

The equation is given by:

y-y_1=m(x-x_1)

We substitute the point and slope into the formula to get;

y-6= - 11(x- - 2)

y= - 11x - 22 + 6

y= - 11x - 16

The correct answer is D


6.which trigonometric function is equivalent to f(x)=sin x? a) f(x)=cos(-x+(3pi/2)) b) f(x)=cos(x+(
Mathematics
1. in abc, c is a right angle and bc= 11. if the measure of angle b= 30degrees, find ac. a) (11sqrt3)/3 b) 3sqrt/11 c)11/2 d)11sqrt3/2 2.which of the following angles has a reference angle of pi/8? a)-51pi/8 b)11pi/8 c)-33pi/8 d)27pi/8 3.your town’s public library is building a new wheelchair ramp to its entrance. by law, the maximum angle of incline for the ramp is 4.76 degrees. the ramp will have a vertical ride of 2.5 ft. what is the shortest horizontal distance that the ramp can span? a)11.9ft b)30.0ft c)4.3ft d)19.1ft 4. find the exact value
Step-by-step answer
P Answered by Specialist

1. The given triangle ABC, has a right angle at C, BC=11, and B=30\degree

\tan 30\degree=\frac{AC}{11}

AC=11\tan 30\degree

AC=\frac{11\sqrt{3}}{3}

Ans: A

2. The reference angle is the angle the terminal side makes with x-axis.

-\frac{33\pi}{8}=-4\frac{\pi}{8}

This implies that, -\frac{33\pi}{8} has a reference angle of \frac{\pi}{8}.

Ans: C

3. Let x be the shortest distance the ramp can span.

From the diagram; \tan (4.76\degree)=\frac{2.5}{x}

\implies x=\frac{2.5}{\tan (4.76\degree)}

\implies x=30.0ft

Ans:B

4. Use the Pythagorean identity: 1+\tan ^2 \theta=\sec^2 \theta.

If \cot \theta=-\frac{1}{2},then  \tan \theta=-2

\implies 1+2^2=\sec^2 \theta

\implies \sec^2 \theta=5

\implies \sec \theta=-\sqrt{5}, In QII, the secant ratio is negative.

Ans:C

5. We have \sin \frac{2\pi}{3}=\frac{\sqrt{3} }{2}

\cos \frac{\pi}{6}=\frac{\sqrt{3} }{2}

\cos \frac{\pi}{3}=\frac{1}{2}

\sin \frac{5\pi}{3}=-\frac{\sqrt{3} }{2}

\cos \frac{7\pi}{6}=-\frac{\sqrt{3} }{2}

\cos \frac{11\pi}{6}=\frac{\sqrt{3} }{2}

Ans:A and D

6.  The given function that is equivalent to f(x)=\sin x is f(x)=\cos (-x+\frac{\pi}{2}).

When we reflect the graph of  f(x)=\cos (x)  in the y-axis and shift it to the left by  \frac{\pi}{2} units, it coincides with graph of f(x)=\sin x.

Ans:C

7. The function y=\tan x is a one-to-one function on the interval [-\frac{\pi}{2},\frac{\pi}{2}]

When we restrict the domain of  y=\tan x on [-\frac{\pi}{2},\frac{\pi}{2}] it becomes an invertible function.

Ans: C

8. The given function is y=3\sin(4x-\pi)

The horizontal shift is given by \frac{C}{B}=\frac{\pi}{4}

The direction of the shift is to the right.

Ans:D

9.  \cos(-75\degree)=\cos(75\degree) by the symmetric property of even functions.

\cos(75\degree)=\cos(45\degree+30\degree)

\cos(75\degree)=\cos(45\degree) \cos30\degree-\sin(45\degree) \sin30\degree

\cos(75\degree)=\frac{\sqrt{2} }{2} \times \frac{\sqrt{3} }{2} -\frac{\sqrt{2} }{2} \times \frac{1}{2}

\cos(75\degree)=\frac{\sqrt{6}-\sqrt{2}}{4}

Ans: B

10. Recall the cosine rule: a^2=b^2+c^2-2bc\cos A

Let the angle measure opposite to the longest side be A, then a=19,b=17, and c=15.

\Rightarrow 19^2=17^2+15^2-2(17)(15)\cos A

\implies -153=-510\cos A

\implies \cos A=0.3

\implies A=\cos^{-1}(0.3)=73\degree

Ans:B

11.  We want to solve 2\sin(2x)\cos(x)-\sin(2x)=0 on the interval;

[-\frac{\pi}{2},\frac{\pi}{2}]

Factor:  \sin2(x)[2\cos(x)-1)=0

Either \sin(2x)=0 \implies x=0\frac{\pi}{2}

Or [2\cos x-1=0 This means that x=\frac{\pi}{3},-\frac{\pi}{3}

Therefore required solution is x=-\frac{\pi}{3},0,\frac{\pi}{3},\frac{\pi}{2}

Ans:D

12. Use the relation:r=\sqrt{x^2+y^2} and \theta=\tan^{-1}(\frac{y}{x})=

The given rectangular coordinate is (1,-2)

This implies that:r=\sqrt{1^2+(-2)^2}=\sqrt{5}

\theta=\tan^{-1}(\frac{-2}{1})= This means  \theta=116.6 or \theta=296.6

The polar forms are: -\sqrt{5},116.6 and \sqrt{5},296.6

Ans: B and C

13.  The polar equation that represents an ellipse is

r=\frac{2}{2-\sin \theta}.

When written in standard form; r=\frac{1}{1-0.5\sin \theta}.

The eccentricity is 0.5\:.

Therefore the r=\frac{2}{2-\sin \theta} is an ellipse.

Ans: B

14. The DeMoivre’s Theorem states that;

(\cos \theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta

This implies that:

[2(\cos \frac{\pi}{9}+i\sin \frac{\pi}{9})]^3=2^3\cos 3\times \frac{\pi}{9}+i\sin 3\times \frac{\pi}{9})

[2(\cos \frac{\pi}{9}+i\sin \frac{\pi}{9})]^3=8(frac{2}{2})+i8(\frac{\sqrt{3}}{2})=4+4\sqrt{3}i

Ans: A

15. Let the initial point be (x,y), Then |v|=\sqrt{(-2-x)^2+(4-y)^2}.

If x=-8, and y=-4.

Then, |v|=\sqrt{(-2--8)^2+(4--4)^2}.

|v|=\sqrt{(-6)^2+(8)^2}=\sqrt{100}=10.

Ans: B

16. We find the dot product to see if it is zero.

u\bullet v=-6(7)+4(10)=-2

Since the dot product is not zero the vectors are not orthogonal

\theta=\cos ^{-1}(\frac{u\bullet v}{|u||v|})

\theta=\cos ^{-1}(-\frac{2}{2\sqrt{13}\times \sqrt{1149} }) =91.3\degree

Ans:B

17. Given v=5i+4j, w=2i-3j.

u=v+w

Add corresponding components

This implies u=(5i+4j)+(2i-3j)

u=(5i+2i+4j-3j)

u=7i+j

Ans:B

See attachment.


1. in abc, c is a right angle and bc= 11. if the measure of angle b= 30degrees, find ac. a) (11sqrt3
1. in abc, c is a right angle and bc= 11. if the measure of angle b= 30degrees, find ac. a) (11sqrt3
Mathematics
Use a trigonometric ratio to find the measure of 0 in the triangle below. Give your answer to the nearest degree. You may use the trigonometric functions of the Desmos scientific calculator:
Step-by-step answer
P Answered by Master

\\ \sf\longmapsto sin\Theta=\dfrac{Perpendicular}{Hypotenuse}

\\ \sf\longmapsto sin\Theta=\dfrac{9}{10}

\\ \sf\longmapsto sin\Theta=0.9

\\ \sf\longmapsto sin^{-1}\Theta=sin^{-1}(0.9)

\\ \sf\longmapsto \Theta=64.15°

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