Answer:
Step-by-step explanation:

<h3><u>Given </u><u>:</u><u>-</u></h3>
- A marker in the center of the fairway is 150 yards away from the centre of the green
- While standing on the marker and facing the green, the golfer turns 100° towards his ball
- Then he peces off 30 yards to his ball
<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>
- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>distance </u><u>between </u><u>the </u><u>golf </u><u>ball </u><u>and </u><u>the </u><u>center </u><u>of </u><u>the </u><u>green </u><u>.</u>
<h3><u>Let's </u><u> </u><u>Begin </u><u>:</u><u>-</u></h3>
Let assume that the distance between the golf ball and central of green is x
<u>Here</u><u>, </u>
- Distance between marker and centre of green is 150 yards
- <u>That </u><u>is</u><u>, </u>Height = 150 yards
- For facing the green , The golfer turns 100° towards his ball
- <u>That </u><u>is</u><u>, </u>Angle = 100°
- The golfer peces off 30 yards to his ball
- <u>That </u><u>is</u><u>, </u>Base = 30 yards
<u>According </u><u>to </u><u>the </u><u>law </u><u>of </u><u>cosine </u><u>:</u><u>-</u>

- Here, a = perpendicular height
- b = base
- c = hypotenuse
- cos theta = Angle of cosine
<u>So</u><u>, </u><u> </u><u>For </u><u>Hypotenuse </u><u>law </u><u>of </u><u>cosine </u><u>will </u><u>be </u><u>:</u><u>-</u>

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>






Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t