All categories

3 years ago

In ΔEFG, f = 42 inches, e = 39 inches and ∠E=43°. Find all possible values of ∠F, to the nearest degree.\

Mathematics

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

The angle F = 47.25°

Step-by-step explanation:

Step(i):-

Given that f =42inches and e = 39 inches

Given that ∠E = 43°

By using sine rule

$\frac{a}{sinA} = \frac{b}{sinB} = \frac{C}{sinC} = 2R$

$\frac{e}{sinE} = \frac{f}{sinF}$

⇒ $\frac{39}{sin43} = \frac{42}{sinF}$

Cross multiplication , we get

⇒ 39 sin F = 42 × sin 43°

$sin F = \frac{42 X 0.68199}{39}$

⇒ sin F = 0.734459

Step(ii):-

⇒ sin⁻¹ (sin F)= sin ⁻¹ ( 0.734459)

⇒ F = 47.25

The angle F = 47.25°

You might be interested in

(GRAPH BELOW)

Rainbow [258]

Answer:

The tap drains approximately half the water from the tank in 15 minutes

The tank initially has 50 gallons of water

Step-by-step explanation:

The tap drains approximately half the water from the tank in 15 minutes. True.

This is true because see that the point (15,25) lies on the graph of the line.

Which means after 15 minutes the amount of water left in the tank is 25 gallons and 25 is half of 50.

The tap drains exactly one gallon from the tank every minute. False

This is false because the slope of the graph is $m=\frac{50-25}{0-15}=-\frac{5}{3}$

The tank initially has 30 gallons of water. False.

The y-intercept is the initial gallons of water which is 50.

The tank initially has 50 gallons of water. True.

The y-intercept is the initial gallons of water which is 50.

The tank takes 50 minutes to drain completely. False.

It is false because the x-intercept tells us the tank is drained completely and it is not 50 minutes but rather 30 minutes.

4 0

3 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$