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oksano4ka [1.4K]
2 years ago
15

A flagpole stands atop a building. From a point 10' away from the base of a building, the angles of elevation of the top and bot

tom of the flagpole measure 80 degrees and 79 degrees respectively. How tall is the building? How long is the flagpole?
(Mostly wondering how to model the problem.)
Mathematics
1 answer:
Debora [2.8K]2 years ago
4 0

Answer:

  building: 51.4 ft

  flagpole: 5.3 ft

Step-by-step explanation:

The relevant trig relation is ...

  Tan = Opposite/Adjacent

If we let b and p represent the heights of the building and flagpole, respectively, then we can write two equations using the tangent relation:

  tan(79°) = b/10

  tan(80°) = (b +p)/10

Multiplying these equations by 10 gives the values we're interested in.

  b = 10·tan(79°) ≈ 51.4 . . . feet

  b +p = 10·tan(80°) ≈ 56.7 . . . feet

Then the height of the flagpole is ...

  p = (b+p) -b = (56.7 ft) -(51.4 ft) = 5.3 ft

The building is 51.4 ft tall.

The flagpole is 5.3 ft tall.

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\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})

We have been given two vectors $\vec{a}$ and $\vec{b}$, we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

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$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

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To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

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