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Gwar [14]
3 years ago
13

Finding angles June 23.2020

Mathematics
1 answer:
finlep [7]3 years ago
3 0

Answer:

angle B is 38 degree

Step-by-step explanation:

In triangle ADC,

finding angle C

angle A + angle C = 110 degree (sum of two interior opposite angles are equal to the exterior angle formed)

46 + angle C = 110

angle C = 110 - 46

angle C = 64 degree

In triangle ABC

finding angle B

angle A + angle B + angle C = 180 degree (sum of interior angles of a triangle)

78 + angle B + 64 = 180

142 + angle B = 180

angle B = 180 - 142

angle B = 38 degree

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zavuch27 [327]

Answer:

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8 0
2 years ago
Factor 3/4 out of 3/4z+6.
Bingel [31]

Answer:

Step-by-step explanation:

5 0
3 years ago
A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of
Gwar [14]

\bold{\huge{\underline{ Solution}}}

<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>

\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}

  • 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>

\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}

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

\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}

\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}

\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}

\sf{ x^{2} = 22500 + 900 + 156.6}

\sf{ x^{2} = 23556.6}

\bold{ x = 153.48\: yards }

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

5 0
2 years ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

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

3 0
3 years ago
Which variable will you choose to eliminate? How? -3x -8y=-24 and <br> 3x - 5y= 45
Jet001 [13]

Answer:

Lets choose y

  • -8y=3x-24
  • y=-3/8x+3

Plug in

  • 3x+15/8x-15=45
  • 39/8x= 60
  • x=480/39
3 0
3 years ago
Read 2 more answers
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