The number of hours of daylight, H, on day t of any given year (on January 1, tequals1) in a particular city can be modeled by the function H(t)equals12 plus 8.3 sine [StartFraction 2 pi Over 365 EndFraction (t minus 83 )]. a. March 24, the 83rd day of the year, is the spring equinox. Find the number of hours of daylight in the city on this day. b. June 24, the 175th day of the year, is the summer solstice, the day with the maximum number of hours of daylight. Find the number of hours of daylight in the city on this day. c. December 24, the 358th day of the year, is the winter solstice, the day with the minimum number of hours of daylight. Find the number of hours of daylight in the city on this day.

The number of hours of daylight, H, on day t of any given year (on January 1, t=1) in a particular city can be modeled by the function:[tex]H(t)=12+8.3 sine \frac{2 \pi}{365}(t-83)[/tex].

(a)On March 24, the 83rd day

t=83

Therefore, the number of hours

[tex]H(t)=12+8.3 sine [\frac{2 \pi}{365}(83-83)][/tex].

H(t)=12 Hours

(b)June 24, the 175th day.

t=175

[tex]H(175)=12+8.3 sine \frac{2 \pi}{365}(175-83)[/tex].

[tex]=12+8.3 sine [\frac{2 \pi}{365}(92)][/tex]=12.23 Hours

(c)December 24, the 358th day

t=358

[tex]H(358)=12+8.3 sine [\frac{2 \pi}{365}(358-83)][/tex].

[tex]=12+8.3 sine [\frac{2 \pi}{365}(275)][/tex]=12.68 Hours

MrRoyal MrRoyal · Answer 2 · 2022-01-13T18:21:08+01:00

A sinusoidal function is one of the basic trigonometric functions

There are 12 hours of daylight on March 24
There are 20.3 hours of daylight on June 24
There are 3.7 hours of daylight on December 24

The function is given as:

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(t - 83))[/tex]

(a) The number of daylight hours on March 24

March 24 is the 83rd day of the year.

So, we have:

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(83 - 83))[/tex]

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(0))[/tex]

Expand the bracket

[tex]H(t) = 12 + 8.3\sin(0)[/tex]

Evaluate sin(0)

[tex]H(t) = 12 + 8.3 \times 0[/tex]

[tex]H(t) = 12 + 0[/tex]

[tex]H(t) = 12[/tex]

Hence, there are 12 hours of daylight on March 24

(b) The number of daylight hours on June 24

June 24 is the 175th day of the year.

So, we have:

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(175 - 83))[/tex]

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(92))[/tex]

Expand the bracket

[tex]H(t) = 12 + 8.3\sin(0.5\pi)[/tex]

Evaluate sin(0.5 pi)

[tex]H(t) = 12 + 8.3\times 1[/tex]

[tex]H(t) = 12 + 8.3[/tex]

[tex]H(t) = 20.3[/tex]

Hence, there are 20.3 hours of daylight on June 24

(c) The number of daylight hours on June 24

December 24 is the 358th day of the year.

So, we have:

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(358 - 83))[/tex]

[tex]H(t) = 12 + 8.3\sin(\frac{2\pi}{365}(275))[/tex]

Expand the bracket

[tex]H(t) = 12 + 8.3\sin(1.5\pi)[/tex]

Evaluate sin(1.5 pi)

[tex]H(t) = 12 - 8.3\times 1[/tex]

[tex]H(t) = 12 - 8.3[/tex]

[tex]H(t) = 3.7[/tex]

Hence, there are 3.7 hours of daylight on December 24

Read more about sinusoidal functions at:

https://brainly.com/question/2410297