Question

In ΔPQR, \overline{PR} PR is extended through point R to point S, \text{m}\angle QRS = (4x-15)^{\circ}m∠QRS=(4x−15) ∘ , \text{m}\angle RPQ = (x+1)^{\circ}m∠RPQ=(x+1) ∘ , and \text{m}\angle PQR = (x-2)^{\circ}m∠PQR=(x−2) ∘ . Find \text{m}\angle RPQ.m∠RPQ.

Answer 1

Answer:

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

Step-by-step explanation:

Answer 2

Answer:

8

Step-by-step explanation:

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