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

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

$q=\frac{p}{3}$

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 $q=\frac{p}{3}$

$q=\frac{90}{3} = 30^{\circ}$

Since, r=p-q

r =$90^{\circ}-30^{\circ}=60^{\circ}$

Answer 2

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=$\frac{1}{3}$p

r is the difference of p and q

r=p-q

As, q=$\frac{p}{3}$

So, r=p-q=p-$\frac{p}{3}$

r=$\frac{3p}{3}-\frac{1p}{3}$

r=$\frac{3p-1p}{3}$

r=$\frac{2p}{3}$

As, p+q+r=180 and q=$\frac{p}{3}$ and r=$\frac{2p}{3}$

So, we get

p+$\frac{p}{3}$+$\frac{2p}{3}$=180

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

3*p+ 3* $\frac{p}{3}$+ 3*$\frac{2p}{3}$=3*180

3p+p+2p=540

6p=540

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

$\frac{6p}{6} =\frac{540}{6}$

p=90

As, p=90

So, q=$\frac{p}{3}$

q=$\frac{90}{3}$

q=30

And, r=$\frac{2p}{3}$

r=$\frac{2*90}{3}$

r=$\frac{180}{3}$

r=60

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

The three angles of triangle are 90,30 and 60