Respuesta :
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
[tex] q=\frac{p}{3} [/tex]
Angle measure of r is the difference of p and q, which implies
[tex] r=p-q [/tex] (Equation 1)
By using the angle sum property of a triangle which states that the sum of all the angles of a triangle is [tex] 180^{\circ} [/tex]
p+q+r=[tex] 180^{\circ} [/tex]
Substituting the value of r from Equation 1,
p+q+p-q=[tex] 180^{\circ} [/tex]
2p=[tex] 180^{\circ} [/tex]
p=[tex] 90^{\circ} [/tex]
Since [tex] q=\frac{p}{3} [/tex]
[tex] q=\frac{90}{3} = 30^{\circ} [/tex]
Since, r=p-q
r =[tex] 90^{\circ}-30^{\circ}=60^{\circ} [/tex]
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=[tex] \frac{1}{3} [/tex]p
r is the difference of p and q
r=p-q
As, q=[tex] \frac{p}{3} [/tex]
So, r=p-q=p-[tex] \frac{p}{3} [/tex]
r=[tex] \frac{3p}{3}-\frac{1p}{3} [/tex]
r=[tex] \frac{3p-1p}{3} [/tex]
r=[tex] \frac{2p}{3} [/tex]
As, p+q+r=180 and q=[tex] \frac{p}{3} [/tex] and r=[tex] \frac{2p}{3} [/tex]
So, we get
p+[tex] \frac{p}{3} [/tex]+[tex] \frac{2p}{3} [/tex]=180
To get rid of fraction, let us multiply the complete equation by 3
3*p+ 3* [tex] \frac{p}{3} [/tex]+ 3*[tex] \frac{2p}{3} [/tex]=3*180
3p+p+2p=540
6p=540
To solve for p, let us divide by 6 on both sides
[tex] \frac{6p}{6} =\frac{540}{6} [/tex]
p=90
As, p=90
So, q=[tex] \frac{p}{3} [/tex]
q=[tex] \frac{90}{3} [/tex]
q=30
And, r=[tex] \frac{2p}{3} [/tex]
r=[tex] \frac{2*90}{3} [/tex]
r=[tex] \frac{180}{3} [/tex]
r=60
So, p=90, q=30, r=60
The three angles of triangle are 90,30 and 60
