Answer:
69.01 m
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you ...
Tan = Opposite/Adjacent
The tangent function is useful for problems like this. Let the height of the spire be represented by h. The distance (d) across the plaza from the first surveyor satisfies the relation ...
tan(50°) = (h -1.65)/d
Rearranging to solve for d, we have ...
d = (h -1.65)/tan(50°)
The distance across the plaza from the second surveyor satisfies the relation ...
tan(30°) = (101.65 -h)/d
Rearranging this, we have ...
d = (101.65 -h)/tan(30°)
Equating these expressions for d, we can solve for h.
(h -1.65)/tan(50°) = (101.65 -h)/tan(30°)
h(1/tan(50°) +1/tan(30°)) = 101.65/tan(30°) +1.65/tan(50°)
We can divide by the coefficient of h and simplify to get ...
h = (101.65·tan(50°) +1.65·tan(30°))/(tan(30°) +tan(50°))
h ≈ 69.0148 ≈ 69.01 . . . . meters
The tip of the spire is 69.01 m above the plaza.
Answer:
Step-by-step explanation:
I can't see the screenshot
so like yeah
Answer:
uhbsu is ysab nq17y 8
Step-by-step explanation:
The length and width of the garden will be obtain as follows:
let the length be x ft, the width will be (x+3.5)ft
area of the bed will be:
A=length*width
11.76=x(x+3.5)
11.76=x^2+3.5x
thus
x^2+3.5x-11.76=0
solving this by quadratic formula we get:
x=[-b+/-sqrt(b^2-4ac)]/2a
thus plugging in the values we obtain
x=[-3.5+/-sqrt(3.5^2-4(-11.76*1))]/2
x=[-3.5+/-sqrt59.29]/2
x=[-3.5+/-7.7]/2
x=4.2 or -11.2
thus the length will be x=4.2 ft and the width will be x-3.5=0.7 ft