Find the measure of ∠c if ∠a = 5 2 x + 30 ∠b = 7 2 x + 40 ∠c = 9 2 x + 10 ∠d = 5 2 x + 20 a) 70° b) 80° c) 90° d) 100°

D) 100°

Step-by-step explanation:

From the question;

∠A = 5 2 x + 30

∠B = 7 2 x + 40

∠C = 9 2 x + 10

∠D = 5 2 x + 20

To solve this question, we first  need to find the value of x and then insert it back in angle c.

The sum of angle in a 4 sided polygon is 360°

∠A + ∠B+ ∠C+  ∠D = 360°

5 2 x + 30  + 7 2 x + 40  +  9 2 x + 10 + 5 2 x + 20  = 360

We can re-arrange the equation

52 x + 72 x + 92 x + 52 x  +  30 + 40 + 10 + 20 = 360

On the left hand-side of the equation, add all the x variables  together and then add all the constants together.

268 x + 100 = 360

subtract 100 from both-side of the equation

268 x + 100 - 100 = 360 - 100

268 x  =  260

Divide both-side of the equation by 268

268x /268  = 260/268

x = 0.97

∠A = 5 2 x + 30  = 52(0.97) + 30 = 50.44 + 30 =80.44≈ 80

∠B = 7 2 x + 40 = 72(0.97) + 40 =69.84 + 40 =109.84≈ 110

∠C = 9 2 x + 10  = 92(0.97) + 10 = 89.24 + 10 =99.24≈ 100

∠D = 5 2 x + 20 = 52(0.97) + 20 = 50.44 + 20 =70.44≈ 70

To check

∠A + ∠B+ ∠C+  ∠D = 360°

80°+ 110° + 100°+70° = 360°

Therefore ∠C =100°

found it xd it ok yt

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl