All categories

3 years ago

If ΔABC is similar to ΔDEF, the m∠A = 50°, and m∠E = 70°, what is m∠C?

Mathematics

2 answers:

siniylev [52]3 years ago

7 0

Answer: A) $60^{\circ}$

Step-by-step explanation:

Given: ΔABC is similar to ΔDEF

Therefore,

∠A=∠D=50°

∠B=∠E=70°

∠C=∠F

Using Angle sum property, in triangle ABC, we get

$\angle{A}+\angle{B}+\angle{C}=180^{\circ}\\\\\Rightarrow\ 50^{\circ}+70^{\circ}+\angle{C}=180^{\circ}\\\\\Rightarrow\ 120^{\circ}\angle{C}+=180^{\circ}\\\\\Rightarrow\ \angle{C}=180^{\circ}-120^{\circ}\\\\\Rightarrow\ \angle{C}=60^{\circ}$

The angle sum property of a triangle says that, the sum of all the angles in a triangle is $180^{\circ}$ .

MAVERICK [17]3 years ago

6 0

Ok so if they are similar that means equal angles so 50+70=120 then 180-120=60 so the answer is A. #TeamAlvaxic

You might be interested in

mel-nik [20]

Answer:

-2

Step-by-step explanation:

maybe?

5 0

3 years ago

Read 2 more answers

Which two grids can be combined to represent 140%?

umka2103 [35]

Can you make the a bit more clear

3 0

4 years ago

Read 2 more answers

Please help I will give you 20 points

zheka24 [161]

You good ?????????????

7 0

3 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

Points (0,0) and (2,1):

y = $\frac{1-0}{2-0}(x-0)$

y = $\frac{x}{2}$

Points (2,1) and (0,3):

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

Mass of the lamina that occupies region D is 4.

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$