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2

## The angles of a teiangle are in ratio 6:5:7.Find the sizes of the three angles?

Jim G.
Featured yesterday

${60}^{\circ} , {50}^{\circ} \text{ and } {70}^{\circ}$

#### Explanation:

$\text{sum the parts of the ratio}$

$\Rightarrow 6 + 5 + 7 = 18$

$\text{the sum of the angles in a triangle } = {180}^{\circ}$

$\Rightarrow {180}^{\circ} / 18 = {10}^{\circ} \leftarrow \textcolor{b l u e}{\text{1 part of the ratio}}$

$\Rightarrow 6 \text{ parts } = 6 \times {10}^{\circ} = {60}^{\circ}$

$\Rightarrow 5 \text{ parts } = 5 \times {10}^{\circ} = {50}^{\circ}$

$\Rightarrow 7 \text{ parts } = 7 \times {10}^{\circ} = {70}^{\circ}$

$\text{the 3 angles are "60^@,50^@" and } {70}^{\circ}$

2

## It has a triangle equal to 180 degrees and I don’t understand this, can you help me?

LucaGasparro
Featured yesterday

See below.

#### Explanation:

Here we are formulating an equation to solve for $x$.

We know that the interior angles of any triangle adds up $180$ degrees.

We have three angles given:

$60$
$x$
$3 x$

This means that :

$60 + 3 x + x = 180$

Now we collect like terms to simplify.

$60 + 4 x = 180$

Now we solve like any linear equation by isolating the variable on one side of the equation with the constant on the other.

Here we must subtract $60$ from both sides to isolate the $x$.

$\therefore 60 + 4 x - 60 = 180 - 60$

$\implies 4 x = 120$
We want one $x$, therefore we divide by the coefficient of $x$ on both sides.

Here we divide by $4$

$4 x = 120$
$\implies x = 30$

We can check if we are right by putting our value of $x$ back into our formulated equation above.

$60 + \left(4 \cdot 30\right) = 60 + 120 = 180$

2

## It has a triangle equal to 180 degrees and I don’t understand this, can you help me?

Hi
Featured yesterday

Triangle sum theorem states that all angles in a triangle must add up to ${180}^{\circ}$, a similar theorem applies to quadrilaterals and it states that all angles in a quad. must add up to ${360}^{\circ}$.

#### Explanation:

You have already applied the triangle sum theorem which states that all 3 angles in a triangle add up to $180$ degrees it seems, so now all you have to do is create an algebraic expression that reflects that.

$\text{Angle 1 + Angle 2 + Angle 3} = {180}^{\circ}$

$x + 3 x + 60 = 180$

$4 x + 60 = 180$

$4 x = 120$

So

$x = {30}^{\circ}$

The angle $x$ is 30 degrees and the angle $3 x$ is $90$ degrees

1

## Two corners of a triangle have angles of  (7 pi )/ 12  and  pi / 4 . If one side of the triangle has a length of  8 , what is the longest possible perimeter of the triangle?

sankarankalyanam
Featured 15 hours ago

Longest possible perimeter of the triangle is

color(blue)(P + a + b + c ~~ 34.7685

#### Explanation:

$\hat{A} = \frac{7 \pi}{12} , \hat{B} = \frac{\pi}{4} , s i \mathrm{de} = 8$

To find the longest possible perimeter of the triangle.

Third angle $\hat{C} = \pi - \frac{7 \pi}{12} - \frac{\pi}{4} = \frac{\pi}{6}$

To get the longest perimeter, smallest angle $\hat{C} = \frac{\pi}{6}$ should correspond to side length 8

Using sine law, $\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

$a = \frac{c \cdot \sin A}{\sin} C = \frac{8 \cdot \sin \left(\frac{7 \pi}{12}\right)}{\sin} \left(\frac{\pi}{6}\right) = 15.4548$

$b = \frac{c \cdot \sin B}{\sin} C = \frac{8 \cdot \sin \left(\frac{\pi}{4}\right)}{\sin} \left(\frac{\pi}{6}\right) = 11.3137$

Longest possible perimeter of the triangle is

color(blue)(P + a + b + c = 15.4548 + 11.3137 + 8 = 34.7685

1

## Two corners of a triangle have angles of  (2 pi )/ 3  and  ( pi ) / 6 . If one side of the triangle has a length of  1 , what is the longest possible perimeter of the triangle?

sankarankalyanam
Featured yesterday

Perimeter of isosceles triangle color(green)(P = a + 2b = 4.464

#### Explanation:

$\hat{A} = \frac{2 \pi}{3} , \hat{B} = \frac{\pi}{6} , s i \mathrm{de} = 1$

To find the longest possible perimeter of the triangle.

Third angle $\hat{C} = \pi - \frac{2 \pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$

It’s an isosceles triangle with
$\hat{B} = \hat{C} = \frac{\pi}{6}$

Least angle $\frac{\pi}{6}$ should correspond to the side 1 to get the longest perimeter.

Applying sine law, $\frac{a}{\sin} A = \frac{c}{\sin} C$

$a = \frac{1 \cdot \sin \left(\frac{2 \pi}{3}\right)}{\sin} \left(\frac{\pi}{6}\right) = \sqrt{3} = 1.732$

Perimeter of isosceles triangle color(green)(P = a + 2b = 1 + (2 * 1.732) = 4.464

1

## A triangle with three degree angle ratio is 4:3:2, this is what the triangle triangle?

sankarankalyanam
Featured yesterday

color(purple)(hatA = 80^@, hatB = 60^@, hatC == 40^@

#### Explanation:

Three angles are in the ratio $4 : 3 : 2$

In a triangle sum of the three angles = ${180}^{\circ}$

Let basic unit x be the unit of each angle.

Then, $\hat{A} = 4 x , \hat{B} = 3 x , \hat{C} = 2 x$

$\therefore 4 x + 3 x + 2 x = {180}^{\circ}$

$9 x = 180$ or $x = {20}^{\circ}$

color(purple)(hatA =) 4 x = 4 * 20 = color(purple)(80^@), color(purple)(hatB = )3x = 3* 20=color(purple)( 60^@), color(purple)(hatC = )2 * x = 2 * 20 = color(purple)(40^@

1

## A cylindrical vessel has radius 3.5cm and height 20cm .It is filled with water. how many metal spheres of radius 3cm can be placed in it.Calculate the volume of water flowing out of the vessel after the it is filled with metal spheres of radius 3cm ?

sankarankalyanam
Featured yesterday

No. Of spheres that the cylinder can contain = 6.

Water overflowing volume = 678.54cc

#### Explanation:

Volume of cylinder ${V}_{c} = \pi {r}^{2} h$

Given : ${r}_{c} = 3.5 c m , {h}_{c} = 20 c m$

V_c = pi (3.5)^2 * 20 = 769.69 cc

Volume of sphere ${V}_{s} = \left(\frac{4}{3}\right) \pi {r}^{3}$

Given r_s= 3 cm V_s = (4/3) pi 3^3 = 113.09cc

No. of spheres that can be placed in the cylinder $N = {V}_{c} / {V}_{s} = \frac{769.69}{113.09} = 6.8$ or only 6 whole spheres can be filled in the cylinder.

Water that will be overflowing after dropping 6 spheres in it = 6 * 113.09 = 678.54 cc

1

## A triangle with three degree angle ratio is 4:3:2, this is what the triangle triangle?

Jim G.
Featured yesterday

${80}^{\circ} , {60}^{\circ} \text{ and } {40}^{\circ}$

#### Explanation:

$\text{sum the parts of the ratio }$

$\Rightarrow 4 + 3 + 2 = 9 \text{ parts}$

• " the sum of the 3 angles in a triangle "=180^@

$\Rightarrow {180}^{\circ} / 9 = {20}^{\circ} \leftarrow \textcolor{b l u e}{\text{1 part}}$

$\Rightarrow 4 \text{ parts } = 4 \times {20}^{\circ} = {80}^{\circ}$

$\Rightarrow 3 \text{ parts } = 3 \times {20}^{\circ} = {60}^{\circ}$

$\Rightarrow 2 \text{ parts } = 2 \times {20}^{\circ} = {40}^{\circ}$

$\text{the 3 angles in the triangle are}$

${80}^{\circ} , {60}^{\circ} \text{ and } {40}^{\circ}$

1

Shiva Prakash M V
Featured yesterday

9)$\text{ x=13.8}$
10)$\text{ x= 8.1}$
11)$\text{ x= 6.0 }$
12)$\text{ x= 5.5}$

#### Explanation:

9)

$B = {37}^{\circ}$
$B C = 11$(adjacent side)
$B A = x$(hypotenuse)
$\cos B = \frac{B C}{B A}$
$\cos {37}^{\circ} = \frac{11}{x}$
$x = \frac{11}{\cos} {37}^{\circ}$
$x = 13.8$

10)

$A = {32}^{\circ}$
$A C = 13$(adjacent)
$C B = x$(opposite)
$\tan A = \frac{C B}{A C}$
$\tan {32}^{\circ} = \frac{x}{13}$
$x = 13 \tan {32}^{\circ}$
$x = 8.1$

11)

$A = {50.1}^{\circ}$
$A C = 5$(adjacent)
$C B = x$(opposite)
$\tan A = \frac{C B}{A C}$
$\tan {50.1}^{\circ} = \frac{x}{5}$
$x = 5 \tan {50.1}^{\circ}$
$x = 6$

12)

$B = {60}^{\circ}$
$B A = 11$(hypotenuse)
$B C = x$(adjacent)
$\cos B = \frac{B C}{B A}$
$\cos {60}^{\circ} = \frac{x}{11}$
$x = 11 \cos {60}^{\circ}$
$x = 5.5$

2

## Please Socratic users help me with this final question?

Jim G.
Featured 15 hours ago

$423.9 \to D$

#### Explanation:

$\text{the volume (V) of a cone is calculated using }$

•color(white)(x)V=1/3pir^2h

$\text{where r is the radius of the base and h the height}$

$\text{we require to find r the radius}$

$\text{using the formula for the circumference (C) of a circle}$

•color(white)(x)C=2pir#

$\Rightarrow 2 \times 3.14 \times r = 56.52$

$\Rightarrow 6.28 r = 56.52$

$\text{divide both sides by } 6.28$

$\frac{\cancel{6.28} r}{\cancel{6.28}} = \frac{56.52}{6.28} \Rightarrow r = 9$

$\Rightarrow V = \frac{3.14 \times 81 \times 5}{3} \approx 423.9 \to D$