Question

The angle measure of thre angle of a triangle are p, q and r. Angle measure of q is one third of p and r is the difference of p and q.

Answer 1

The angle measure of three angle of a triangle are p, q and r.

As, sum of angles of triangle is 180 degrees.

So, p+q+r=180

Angle measure of q is one third of p

q=
`(1)/(3)`p

r is the difference of p and q

r=p-q

As, q=
`(p)/(3)`

So, r=p-q=p-
`(p)/(3)`

r=
`(3p)/(3)-(1p)/(3)`

r=
`(3p-1p)/(3)`

r=
`(2p)/(3)`

As, p+q+r=180 and q=
`(p)/(3)` and r=
`(2p)/(3)`

So, we get

p+
`(p)/(3)`+
`(2p)/(3)`=180

To get rid of fraction, let us multiply the complete equation by 3

3*p+ 3*
`(p)/(3)`+ 3*
`(2p)/(3)`=3*180

3p+p+2p=540

6p=540

To solve for p, let us divide by 6 on both sides

`(6p)/(6) =(540)/(6)`

p=90

As, p=90

So, q=
`(p)/(3)`

q=
`(90)/(3)`

q=30

And, r=
`(2p)/(3)`

r=
`(2*90)/(3)`

r=
`(180)/(3)`

r=60

So, p=90, q=30, r=60

The three angles of triangle are 90,30 and 60

Answer 2

In this question, i think we have to determine all the three angles of a triangle.

Since the three angles of a triangle are p,q and r.

Since, angle measure of q is one third of p, which implies

`q=(p)/(3)`

Angle measure of r is the difference of p and q, which implies

`r=p-q` (Equation 1)

By using the angle sum property of a triangle which states that the sum of all the angles of a triangle is
`180^(\circ)`

p+q+r=
`180^(\circ)`

Substituting the value of r from Equation 1,

p+q+p-q=
`180^(\circ)`

2p=
`180^(\circ)`

p=
`90^(\circ)`

Since
`q=(p)/(3)`

`q=(90)/(3) = 30^(\circ)`

Since, r=p-q

r =
`90^(\circ)-30^(\circ)=60^(\circ)`