When the sun is at a certain angle in the sky, a 50-foot building will cast a -foot shadow. What is the length of the shadow in

Answer:

The height of the triangular pyramid base is represented by s√3/2 units.

Step-by-step explanation:

This question focuses on calculating the height of the equilateral triangle at the base of the oblique pyramid.

According to the question, the equilateral triangle side has a length of a units.

Let’s review some properties of equilateral triangles:

a. All sides are identical in length, in this case, side s represents all sides.

b. All angles measure the same, at 60 degrees each.

c. Drawing a perpendicular line from the top vertex to the base divides the triangle into two right triangles, which have angles of 60 and 30 degrees, respectively.

To find the height of this triangular base, we can utilize either of the two right-angled triangles.

Recall the angles of 30, 60 and the side length s.

For height h calculation, trigonometric functions will assist us.

The relevant trigonometric relationship involves the sine of the angle (where side length s is the hypotenuse and height h is the opposite side relative to the 60-degree angle).

Thus, we can express this as:

Sine of an angle = opposite side length/hypotenuse length

sin 60 = h/s

Consequently, h = s sin 60.

In surd form, this yields:

sin 60 = √3/2.

Therefore, we find:

h = s * √3/2 = s√3/2 units.

Answer:

55°

Step-by-step explanation:

The problem does not have a suitable diagram. Please check the attachment for the diagram.

Observing the diagram, it is evident that the triangle is isosceles, as it has two sides that are congruent (i.e., the same length). Since these sides are identical, the base angles are consequently also equal.

From the diagram, <EFG = <EGF

Since  <EFG  = 55°, it follows that  <EGF = y = 55°

Thus, the value of y is established as 55°

Answer:

Step-by-step explanation:

Step 1:-

We have c1(t) = e^ t i + (sin(t))j + t³k

and c2(t) = e^−t i + (cos(t))j − 6t³k.

By adding c1(t) and c2(t):

c1(t)+c2(t) = e^ t i + (sin(t))j + t³k + e^−t i + (cos(t))j − 6t³k

Now, employing the derivative formula:

Next, differentiate with respect to 't'

By factoring out i, j, and k terms, we arrive at:

The maximum distance visible on Earth is calculated using the formula

= your initial height and = your secondary height

In this case, represents the height of the periscope and denotes the height of the ship, leading us to a distance of 832.45553203368 feet. However, rounding to the nearest mile yields the answer as

If there are any discrepancies, please inform me and I will recalculate!

The diagrams for parts A and C are included here. For part B, we have circle O. We begin by drawing two radii OA and OC, connecting points A and C to create chord AC. The radius intersects chord AC at point B, bisecting AC into equal segments AB and BC. This gives us two triangles, ΔOBA and ΔOBC, where OA equals OC (since they're radii), OB equals OB (by the reflexive property), and AB is equal to BC (as stated in the question). By applying the SSS triangle congruence criterion, we conclude that ΔOBA is congruent to ΔOBC, allowing us to deduce that ∡OBA equals ∡OBC, both measuring 90°. Thus, OB is perpendicular to AC. Moving on to part D, we again work with circle O and draw the two radii OA and OC, joining points A and C to create chord AC. The radius intersects AC at point B, where AB is perpendicular to AC, meaning ∡B equals 90°. We then consider the right triangles ΔOBA and ΔOBC, and given OA equals OC (the radii), and OB equals OB (reflexive property), we conclude through the HL triangle congruence that ΔOBA is congruent to ΔOBC. Consequently, we find BA equal to BC, thus OB bisects AC.