Ask question

Ask question

All categories

1 year ago

9

A straight road makes an angle of 15 degrees with the horizontal. When the angle of elevation of the sun is 57 degrees, a vertic

al pole at the side of the road casts a shadow 75 feet long directly down the road, as shown in the figure. Approximate the length of the pole. Round to the nearest hundredth.

1 answer:

Vilka [71]1 year ago

7 0

The length of the pole is 19.15 feet.

<h3>What is the angle of elevation?</h3>

The angle of elevation is the angle created when an observer looks at an object placed above its height in relation to the eye level or the horizontal line. An angle of elevation is, for example, the angle created between the line of sight and the horizontal line when a man on Earth observes the Sun.

Given:

Angle made by straight road = 15°

Angle of elevation of the sun = 57°

Length of the shadow = 75 feet

∠ACB = 90° - 57° = 33°

∠CAB = 57° - 15° = 42°

Using Sine Law to find BC i.e.length of the pole.

$\frac{BC}{sin A} = \frac{AB}{sin C}$

$\frac{BC}{sin 42} = \frac{75}{sin 33}$

$BC= \frac{75sin 42}{sin 33}$

BC ≅ 19.15 feet (to the nearest hundredth.)

Hence, the approximate length of the pole is 19.15 feet.

To know more about angle of elevation visit:

brainly.com/question/16716174

#SPJ10

You might be interested in

Which expression describes the representation given?

ser-zykov [4K]

Answer:

b.

Step-by-step explanation:

those are a lot of tabs

4 0

3 years ago

Read 2 more answers

two circles intersect each other in A and C. O the centre of the smaller circle, lies on circumference of the bigger circle D is

olchik [2.2K]

Answer:

ss

Step-by-step explanation:

s

5 0

3 years ago

What is the value of h when the function is converted to vertex form?<br> g(x) = x^2 – 6x + 14

madam [21]

Hello,

g(x)=x²-6x+14=x²-2*3x+9+14-9=(x-3)²+5
h=3, k=5

6 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

4 years ago

Can someone tell me the answer to this one?

rewona [7]

Answer:

eeeeeeeeeeeeeeeeeee

Step-by-step explanation:

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

5 0

3 years ago