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3 years ago

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1 answer:

RoseWind [281]3 years ago

7 0

Answer:

The angle of depression from the lighthouse is the angle of elevation from the boat. The line of sight from the top of the lighthouse, when depressed, becomes an alternate interior angle of two parallel lines, the line of sight from the top and the boat's movement.

Step-by-step explanation:

From the boat's standpoint, which is what we want anyway, there is a right triangle, with the distance from the lighthouse the adjacent side, the opposite is 200,' and the boat the point desired. The tangent is needed.

tangent 18.33 degrees (33 minutes is 33/60 of a degree)=200/x

x=200/tan 18.33=592.58 feet

nearer, the angle of elevation increases as the adjacent side gets smaller relative to the opposite side.

tangent 51.65 degrees (33 minutes is 33/60 of a degree)=200/x

x=200/tan 51.65=158.23 feet

That difference is 434.35 feet

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Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

$x^{2} + 3x - 8^{- 14} = 0$

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above solution is for the quadratic equation of the form:

$ax^{2} + bx + c = 0$

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the given eqn

a = 1

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Now, using the above values in the formula mentioned above:

$x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)$

Now, Rationalizing the above eqn:

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

$x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

Solving the above eqn:

$x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}$

Solving with the help of caculator:

$x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}$

The precise value upto three decimal places comes out to be:

$x_{1, 1'} = 0.758\times 10^{- 14}$

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