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3 years ago

How tall is a building that casts a 78-ft shadow when the angle of elevation of the sun is 30 degrees?

Mathematics

1 answer:

Marta_Voda [28]3 years ago

8 0

Answer: $45.03\ ft$

Step-by-step explanation:

Given

The length of the shadow is 78 ft

the angle of elevation is $\theta =30^{\circ}$

Suppose the height of the building is $h$

From the figure, we can write

$\tan 30^{\circ}=\dfrac{h}{78}\\\\h=78\tan 30^{\circ}\\\\h=45.03\ ft$

The height of the building is 45.03 ft

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3 years ago

(08.07 HC)

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Answer:

$\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}$

$\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

C) See attachment.

Step-by-step explanation:

Given function:

$f(x)=2x^2-x-10$

To factor a <u>quadratic</u> in the form $ax^2+bx+c$ <em> , </em>find two numbers that multiply to $ac$ and sum to $b$ :

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Factor out the common term (2x - 5):

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The x-intercepts are when the curve crosses the x-axis, so when y = 0:

$\implies (x+2)(2x-5)=0$

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So the x-intercepts are:

$x=-2, \:\:x=\dfrac{5}{2}$

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$\implies x=\dfrac{-b}{2a}$

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$\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}$

To find the y-value of the vertex, substitute the found value of x into the function:

$\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}$

Therefore, the vertex of the function is:

$\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

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The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).

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How tall is a building that casts a ​78-ft shadow when the angle of elevation of the sun is 30 degrees​?

How tall is a building that casts a 78-ft shadow when the angle of elevation of the sun is 30 degrees?