Question

f the measure of angle BCD is 67.4°, the measure of angle ABC is °. If the length of base AB is 80 feet, the length of DC is feet.

Answer 1

Answer:

$\angle ABC = 112.6^o$

$DE =105$

Step-by-step explanation:

Given

See attachment for complete question

Required

$\angle ABC$ and side length DC

 Calculating $\angle ABC$

First, calculate $\angle EBC$

$\angle EBC + \angle BCE + 90 = 180$ --- angles in a triang;e

$\angle BCE =\angle BCD = 67.4$

So:

$\angle EBC + 67.4+ 90 = 180$

$\angle EBC + 157.4 = 180$

Collect like terms

$\angle EBC = 180-157.4$

$\angle EBC = 22.6$

So:

$\angle ABC = \angle EBC +90$

$\angle ABC = 22.6 +90$

$\angle ABC = 112.6^o$

Calculating DC

First, calculate EC using Pythagoras theorem.

$BC^2 = BE^2 + EC^2$

$65^2 = 60^2 + EC^2$

$4225 = 3600 + EC^2$

Collect like terms

$EC^2 =4225 - 3600$

$EC^2 =625$

Take square roots

$EC = 25$

Length DC is:

$DC = DE + EC$

Where

$DE = AB =80$

So:

$DE =80+25$

$DE =105$

Answer 2

Answer:

ABC= 112.6

DC= 105

Step-by-step explanation:

got it right on edge!