All categories

3 years ago

Solve the triangle. A = 54°, b = 11, c = 8

1 answer:

5 0

In this problem, you apply principles in trigonometry. Since it is not mentioned, you will not assume that the triangle is a special triangle such as the right triangle. Hence, you cannot use Pythagorean formulas. The only equations you can use is the Law of Sines and Law of Cosines.

For finding side a, you can answer this easily by the Law of Cosines. The equation is

a2=b2 +c2 -2bccosA
a2 = 11^2 + 8^2 -2(11)(8)(cos54)
a2 = 81.55
a = √81.55
a = 9

Then, we use the Law of Sines to find angles B and C. The formula would be

a/sinA = b/sinB = c/sinC
9/sin54° = 11/sinB
B = 80.4°

9/sin54° = 8/sinC
C = 45.6°

The answer would be: a ≈ 9, C ≈ 45.6, B ≈ 80.4

You might be interested in

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago

Will you find the volume for 30 points?

morpeh [17]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

William has 8 1/4 cups of fruit juice. If he divides the juice into 3/4 cup servings, how many servings will he have?

nikdorinn [45]

Answer: 11 servings

Step-by-step explanation:

To get the number of servings, all we need to do is divide 8 1/4 by 3/4

converting 8 1/4 to improper fraction will yield 33/4

8 1/4 ÷ 3/4

=33/4 ÷ 3/4

= 33/4 ÷ 4/3

The 4 at the numerator will cancel the 4 at the denominator. we will be left with;

=33/3 =11

Therefore, there will be 11 servings.

7 0

3 years ago

Read 2 more answers

If a 42.9 ft tall flagpole casts a 253.1 ft long shadow then how long is the shadow that a 6.2 ft tall woman casts?

blsea [12.9K]

We will have to use proportions:
$\frac{42.9}{253.1} = \frac{6.2}{x}$
Cross multiplying we get
$42.9x = 6.2 \times 253.1$
Now divide:
$x = \frac{6.2 \times 253.1}{42.9}$
Solve to tenths place we get
$x = 36.6$
The woman casts a 36.6 ft long shadow.

3 0

3 years ago

The table shows the results of an experiment in which subjects taking either a drug or a placebo either reported fatigue or did

Lady bird [3.3K]

Answer:

Therefore, the correct options are;

-P(fatigue) = 0.44

$P(fatigue \left | \right drug)$ = 0.533

--P(drug and fatigue) = 0.32

P(drug)·P(fatigue) = 0.264

Step-by-step explanation:

Here we have that for dependent events,

$P(A \, and \, B)= P(A)\times P(B\left | \right A )$

From the options, we have;

$P(fatigue \left | \right drug)$ = 0.533

P(drug) = 0.6

P(drug and fatigue) = 0.32

Therefore

P(drug and fatigue) = P(drug)× $P(fatigue \left | \right drug)$

= 0.6 × 0.533 = 0.3198 ≈ 0.32 = P(drug and fatigue)

Therefore, the correct options are;

-P(fatigue) = 0.44

$P(fatigue \left | \right drug)$ = 0.533

--P(drug and fatigue) = 0.32

P(drug)·P(fatigue) = 0.264

Since P(fatigue) = 0.44 ∴ P(drug) = 0.264/0.44 = 0.6.

4 0

3 years ago