Trigonometry?

Question

When the suns angle of elevation is 30° the shadow of a post is 6ft longer than when the angle is 45°. Find the height of the post.

if you could explain it to me step by step,it would be very helpful.

Answer 1

This is a case of using triangles and the tangent function. Let's imagine the sun at 30° above the horizon and the shadow a pole would cast due to it. The sun makes a triangler with the angle being 30°. Also, the pole casts a shadow. If you draw a line from the end of the shadow to the top of the pole, the angle is also 30°. The length of the shadow is 6 + x, where x is the length when the sun is at 45°. The height of the pole is y.

Now, when the sun is at 45°, the length of the shadow is x. Because the triangle formed by the shadow and the pole is a 45-45-90 triangle, the shadow and pole height must be the same, or both are x.

Now, we know that y=x. We have a triangle with an angle of 30°, the opposite side with a length of x, and the adjacent side with a length of x+6. We can now solve:

tan 30° = x/(x+6)
(x+6)tan 30° = x
(1-tan 30°)x = 6tan 30°
x = 6tan 30° / (1-tan 30°)
x = 8.2 feet

Answer 2

8 feet (or 8.20 feet, if preferred)

Consider the triangles made with the sun rays being the hypotenuse, the post being the height, and the shadow length being the base. When the sun is at 45 degrees, the height of the post is the opposite side to the 45 degree angle and the base (which I will call x) is the adjacent side so the ratio of the post height to the base equals the tan 45 dgrees: post / base = tan 45 = post / x. Similarly, when the sun is at 30 degrees, we find this relationship (remembering that the shadow (the base) is 6 feet longer: x + 6): post / (x + 6) = tan 30.

So, if we solve each for "post" we get: x * tan 45 = (x + 6) * tan 30 = post. Solve the first two parts for x: x * tan 45 = x * tan 30 + 6 * tan 30 and x * (tan 45 - tan 30) = 6 * tan 30 so: x = (6 * tan 30) / (tan 45 - tan 30).

Substitute back into the first equation after isolating "post": post = tan 45 * { (6 * tan 30) / (tan 45 - tan 30) } and solve the right hand side with whatever means you find convenient to get the post height = 8.196152 feet. Properly speaking, it should be rounded to 1 significant digit, but I'll leave it at 8.20 feet.

Answer 3

Let the top of the post be A, the bottom of the post be B and the edge of the shadow be C.

ABC is a right triangle and ABC is the right angle.
BCA is the angle of the sun's elevation

Suppose the length of the post is p

The tangent of angle BCA is the opposite side (the post) divided by the adjacent side (the shadow) and

if the sun's elevation is 30 then tan(30) is √3 / 3

Since the shadow is p+6 feet, and the tangent is p/(p+6) (opposite over adjacent) then p/(p+6) = √3/ 3 and we can solve for p

3p = √3p + √36

(3-√3)p = √36

(√3-1)p = 6

p = 6/(√3-1)

p = 6/(√3-1) * (√3+1)/(√3+1)

p = 6(√3+1)/2

p = 3(√3+1)

Answer 4

Draw a right angled triangle, height (h) and base (x), make the angle 45 degrees

Now superimpose a second triangle extended because the angle is 30 degrees, the distance between these angles (along the base) is 6

Therefore the total base is x+6

We can now use trigonometry to say:

Tan (45) = h / x

Rearranged to get:

x = h / Tan (45)

AND

Tan 30 = h / (x+6)

Rearranged to get:

h = x Tan (30) + 6 Tan (30)

Now substitute in for x from equation one to get:

(h / Tan 45 ) Tan (30) + 6 Tan (30) = h

This is quickly solved to get:

h = 8.196 feet

Answer 5

Draw two similar triangles. They both have the same 90° angle. They both have the height of the post as one leg, which I will label x. The longer triangle has angle 90° to the floor, and side L2 along the floor. The shorter triangle has angle 45° to the floor, and side L1 along the floor. Draw L2 bigger than L1.

L2 = 6 + L1
tan 45 = x / L1
tan 30 = x / L2
3 equations in 3 unknowns
x = 6 tan30 tan45 / (tan 45 - tan 30)
Answer: 8.2 ft.

Answer 6

Draw yourself a 45 - 45 right triangle with the post being the vertical line. Call the vertical line y and the horizontal line x.

Now, extend the the horizontal line a little bit and call that length 6ft. Next, draw a diag. line to the top of the vertical line. Assume that this new angle is 30 degrees.

We know that:

1) tan45 = 1 = y/x and
2) tan30 = 0.577 = y / (x + 6)

using 1), we see that y = x, and multiplying the second equation by (x+6), we get:

0.577(x+6) = y now plug in y for x from 1) and we get:

0.577y + 3.4638 = y

solve and we get y = 8.2ft

Answer 7

Have two right angled as follows:-
Triangle 1
vertical = h = height of post.
base angle = 45°
shadow length = x ft
tan 45° = h/x
h = x.tan 45°

Triangle 2
vertical height = h = height of post.
base angle = 30°
shadow length = x + 6 ft
tan 30° = h / (x + 6)
h = (x + 6).tan 30°

x.tan 45° = (x + 6).tan 30°
x = (x + 6).(1/√3)
√3.x = x + 6
0.732 x = 6
x = 6 / 0.732
x = 8.2

tan 45° = h / 8.2
h = 8.2 ft

Answer 8

Let O be the base of the post, C the top.
Let A and B be hte ends of the 45 adn 30 deg shadow resp.

So in triangle CAB has AB = 6, angle A = 135, C = 15, B = 30.
Using the sine rule, BC = 6*sin135 / sin15

Now in triangle OBC, OC = BC*sin30

You'll get OC = 8.20 feet.

Answer 9

Let y be the height of the post
Let x be the shadow at 45 degrees
Then tan 30 = y/(x+6)
and tan 45 = y/x or x = y/(tan 45)
Now use the value of x in the first eqn:
from the first eqn y = (x+6)tan 30
and then y = [(y/tan 45) + 6]tan 30
Now just solve for y and you are home in time for ice cream

Answer 10

Tan(45) = H/x

Tan(30) = H / ( x + 6)


solve for H
H = tan(45)x
H = tan(30)(x + 6)

tan(45)x = tan(30)(x + 6)
x = tan(30)x + tan(30)6
x - tan(30x) = tan(30)6
use a calculator an you'll get x = 8.2ft

H = 8.2 (tan(45))
H = 8.2ft

hope this helps