53Find the x of the angle

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

[tex]\tan 30\degree=\frac{AC}{11}[/tex]

[tex]AC=11\tan 30\degree[/tex]

[tex]AC=\frac{11\sqrt{3}}{3}[/tex]

Ans: A

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

[tex]-\frac{33\pi}{8}=-4\frac{\pi}{8}[/tex]

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

Ans: C

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

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

[tex]\implies x=\frac{2.5}{\tan (4.76\degree)}[/tex]

[tex]\implies x=30.0ft[/tex]

Ans:B

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

If [tex]\cot \theta=-\frac{1}{2}[/tex],then  [tex]\tan \theta=-2[/tex]

[tex]\implies 1+2^2=\sec^2 \theta[/tex]

[tex]\implies \sec^2 \theta=5[/tex]

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

Ans:C

5. We have [tex]\sin \frac{2\pi}{3}=\frac{\sqrt{3} }{2}[/tex]

[tex]\cos \frac{\pi}{6}=\frac{\sqrt{3} }{2}[/tex]

[tex]\cos \frac{\pi}{3}=\frac{1}{2}[/tex]

[tex]\sin \frac{5\pi}{3}=-\frac{\sqrt{3} }{2}[/tex]

[tex]\cos \frac{7\pi}{6}=-\frac{\sqrt{3} }{2}[/tex]

[tex]\cos \frac{11\pi}{6}=\frac{\sqrt{3} }{2}[/tex]

Ans:A and D

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

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

Ans:C

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

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

Ans: C

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

The horizontal shift is given by [tex]\frac{C}{B}=\frac{\pi}{4}[/tex]

The direction of the shift is to the right.

Ans:D

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

[tex]\cos(75\degree)=\cos(45\degree+30\degree)[/tex]

[tex]\cos(75\degree)=\cos(45\degree) \cos30\degree-\sin(45\degree) \sin30\degree[/tex]

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

[tex]\cos(75\degree)=\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]

Ans: B

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

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

[tex]\Rightarrow 19^2=17^2+15^2-2(17)(15)\cos A[/tex]

[tex]\implies -153=-510\cos A[/tex]

[tex]\implies \cos A=0.3[/tex]

[tex]\implies A=\cos^{-1}(0.3)=73\degree[/tex]

Ans:B

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

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

Factor:  [tex]\sin2(x)[2\cos(x)-1)=0[/tex]

Either [tex]\sin(2x)=0 \implies x=0\frac{\pi}{2}[/tex]

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

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

Ans:D

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

The given rectangular coordinate is (1,-2)

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

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

The polar forms are: [tex]-\sqrt{5},116.6[/tex] and [tex]\sqrt{5},296.6[/tex]

Ans: B and C

13.  The polar equation that represents an ellipse is

[tex]r=\frac{2}{2-\sin \theta}[/tex].

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

The eccentricity is [tex]0.5\:[/tex].

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

Ans: B

14. The DeMoivre’s Theorem states that;

[tex](\cos \theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta[/tex]

This implies that:

[tex][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})[/tex]

[tex][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[/tex]

Ans: A

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

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

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

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

Ans: B

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

[tex]u\bullet v=-6(7)+4(10)=-2[/tex]

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

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

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

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.

a,c,d,a,b,a,a,c,b,

step-by-step explanation:

do this

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

Ans: A

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

This implies that,  has a reference angle of .

Ans: C

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

From the diagram;

Ans:B

4. Use the Pythagorean identity: .

If ,then  

, In QII, the secant ratio is negative.

Ans:C

5. We have

Ans:A and D

6.  The given function that is equivalent to  is .

When we reflect the graph of    in the y-axis and shift it to the left by   units, it coincides with graph of .

Ans:C

7. The function  is a one-to-one function on the interval

When we restrict the domain of   on  it becomes an invertible function.

Ans: C

8. The given function is

The horizontal shift is given by

The direction of the shift is to the right.

Ans:D

9.   by the symmetric property of even functions.

Ans: B

10. Recall the cosine rule:

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

Ans:B

11.  We want to solve  on the interval;

Factor:  

Either

Or  This means that

Therefore required solution is

Ans:D

12. Use the relation: and

The given rectangular coordinate is (1,-2)

This implies that:

This means   or

The polar forms are:  and

Ans: B and C

13.  The polar equation that represents an ellipse is

.

When written in standard form; .

The eccentricity is .

Therefore the  is an ellipse.

Ans: B

14. The DeMoivre’s Theorem states that;

This implies that:

Ans: A

15. Let the initial point be (x,y), Then .

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

Then, .

.

Ans: B

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

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

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.

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Igot a c d a b a a c b

74 degrees

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74 degrees

Step-by-step explanation:

These two triangles are isosceles.

This means that the top angle of the left triangle is also 53 degrees.

All triangles have 180 degrees.

When we subtract 2(53) from 180 we get 74.

We can use this vertical angle to find out the angle left of x which is 74.

Since the second triangle is also isosceles, x is equal to 74 degrees.