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.

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:

60 boxes

Step-by-step explanation:

Volume of godown = l*b*h

=(72*60*24)m³

Volume of 1 box = (12*12*12)m³

Number of boxes that can be fit in godown= volume of godown/volume of 1 box

= (72*60*24)m³/(12*12*12)m³

= 103680/1728

= 60

Hope it helps