What would be the angle of elevation of a tree from the ground, if the height of the

Mathematics

1 answer:

garri49 [273]3 years ago

7 0

Answer:

45°

Step-by-step explanation:

The diagram for this question has been attached to this response. Please check.

The angle of elevation is the angle between a horizontal line from a viewer and the line of sight to an object being viewed which is above the horizontal line.

From the diagram;

θ is the angle of elevation

x = height of the tree

y = length of the shadow of the tree = x

Therefore,

tanθ = $\frac{x}{y}$ [Remember that y = x? Then substitute into the equation]

tanθ = $\frac{x}{x}$

tanθ = 1

θ = tan⁻¹(1)

θ = 45°

Therefore, the angle of elevation is 45°

You might be interested in

Consider the force field and circle defined below.

grin007 [14]

By Green's theorem,

$\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy$

where $C$ is the circle $x^2+y^2=9$ and $D$ is the interior of $C$ , or the disk $x^2+y^2\le1$ .

Convert to polar coordinates, taking

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

Then the work done by $\vec F$ on the particle is

$\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0$