An isosceles triangle have two equal sides and two equal angles.
The measure of angles in [tex]\triangle PQR[/tex] are: 20, 20 and 140 degrees
Given that:
[tex]PQ = PR = 45mm[/tex]
[tex]\angle PQR = 20^o[/tex]
I've added as an attachment, triangle [tex]\triangle PQR[/tex]
From the attached triangle, we have:
[tex]\angle PQR = \angle QRP = 20^o[/tex] --- base angles of isosceles triangle
So, we have:
[tex]\angle PQR + \angle QRP +\angle RPQ = 180^o[/tex] --- angles in a triangle
This gives:
[tex]20^o + 20^o +\angle RPQ = 180^o[/tex]
[tex]40^o +\angle RPQ = 180^o[/tex]
Collect like terms
[tex]\angle RPQ = 180^o - 40^o[/tex]
[tex]\angle RPQ = 140^o[/tex]
Hence, the angles of [tex]\triangle PQR[/tex] are:
[tex]\angle RPQ = 140^o[/tex]
[tex]\angle PQR = 20^o[/tex]
[tex]\angle QRP = 20^o[/tex]
Read more about isosceles triangles at:
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