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.
<u>Step-by-step explanation:</u>
Here we have , cos x = -4/5 ( corrected as given -45 which is not possible ) ,
180° < x < 270° i.e. in third quadrant . We need to find tan(2x) . Let's find out:
⇒
⇒
⇒
⇒
⇒
⇒
⇒ , since in third quadrant .
remains same .
Now ,
⇒
⇒
⇒
⇒
Therefore, .