In δpqr,

pr

is extended through point r to point s, m∠qrs =(10x−12)∘(10x−12) ∘ , m∠pqr = (3x+20)∘(3x+20) ∘ , and m∠rpq=(3x−8)∘(3x−8) ∘ . find m∠qrs.

Step-by-step explanation:

Notice that, the angle QRS is external to the triangle and adjacent to the angle PRQ. According to the theorem of a external/adjacent angle, we have: m∠QRS = m∠PQR + m∠RPQ, where PQR and RPQ are internal angles.

From the hypothesis, we have:

m∠QRS =(10x−12)∘(10x−12)

m∠PQR = (3x+20)∘(3x+20)

m∠RPQ=(3x−8)∘(3x−8)

Using the first equation and replacing the hypothesis:

m∠QRS = m∠PQR + m∠RPQ

(10x−12)∘(10x−12) = (3x+20)∘(3x+20) + (3x−8)∘(3x−8)

Multiplying and applying the remarkable identity:

Then, we use a calculator to find the roots, which are:

In this case, we will see what root is the right one.

Now, we replace it into m∠QRS =(10x−12)∘(10x−12), because we need to find m∠QRS.

m∠QRS =(10x−12)∘(10x−12) = (10(4.7) - 12) (10(4.7) - 12) = (35) (35) = 1225

100.5°

Step-by-step explanation:

Data

m∠QRS = (10x−12)°

m∠PQR = (3x+20)°

m∠RPQ = (3x−8)°

The three angles form the triangle ΔPQR, then the addition of them makes 180°.

m∠QRS + m∠PQR  + m∠RPQ  = 180°

10x−12 + 3x+20 + 3x−8  = 180

16x = 180

x = 180/16 = 11.25

Then, m∠QRS = 10*11.25−12 = 100.5°

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