All categories

2 years ago

the ratio of the angle measures of a triangle are 3:3:4 what angle measures the triangles classification

Mathematics

1 answer:

TEA [102]2 years ago

6 0

Step-by-step explanation:

sum of the ratio=3+3+4=10

now,

angle I = 3/10×180=54

angle II= 3/10×180=54

angle III=4/10×180=72

that shows th given triangle is isoceles

You might be interested in

The radius of a cylinder is 3 cm and the height is 6 cm.

Ratling [72]

Answer

Step-by-step explanation:

A=2πrh+2πr2=2·π·3·6+2·π·32≈169.646

5 0

2 years ago

Read 2 more answers

What is the area of an equilateral triangle that has a perimeter of 36 centimeters? Round to the nearest square centimeter.

Vika [28.1K]

Answer:

(a is size of side of triangle)

the perimeter of equilateral triangle =3a

or, 36=3a

Thus, a = 13cm

area of equilateral triangle

=√3/4*(a^2 )

=√3/4*(13^2)

=73cm^2

5 0

2 years ago

In a right triangle, the acute angles measure ×+15 and 2× degrees. What is the measure of the larger acute angle of the triangle

harkovskaia [24]

Sum of angles in a triangle is 180°

90 + (x + 15) + (2x) = 180

90 + x + 15 + 2x = 180

105 + 3x = 180

3x = 180 - 105

3x = 75

x = 75 ÷ 3

x = 25

Smaller angle = x + 15 = 25 + 15 = 40°

Larger angle = 2x = 2(25) = 50°

Answer: 50°

5 0

3 years ago

Two angles of a triangle measure 57 and 46. What is the measure of the largest exterior angle of the triangle?

irina [24]

Answer:

A. 103°

Step-by-step explanation:

57° + 46° =103°

This shows that the third angle in the triangle is 180°-103°=77°

Therefore the largest exterior angle is 180°-77°=103°

5 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

3 years ago

the ratio of the angle measures of a triangle are 3:3:4 what angle measures the triangles classification ​

the ratio of the angle measures of a triangle are 3:3:4 what angle measures the triangles classification