Given: δstq, st = tq line sd is perpendicular to tq m∠1=32° find: m∠s, m∠t, m∠q

 m∠S = 58 ° ;  m∠T = 64 ° ;  m∠Q = 58 ° .


Explanation:

  
To find "m∠Q" :

Consider the triangle formed by ∠D , ∠1, and ∠Q  ; 

  Given m∠1 = 32° ; 
   
       and  m∠D = 90° (a "right triangle" ; as shown in "image attached" ; 

and by definition, all triangles have 3 angles and 3 sides; and all 3 angles of a triangle add up to 180° ; 

→  m∠Q  =  180 − (90 + 32) = 180 − 122 = 58 ; 


Since this is an "isosceles triangle" ;  

→  m∠S = m∠Q = 58° .

Since all angles of any triangle add up to " 180° " ; 

→ m∠T = 180 − (58 + 58) = 180 − 116 = 64 ; 


 m∠S = 58 ° ;  m∠T = 64 ° ;  m∠Q = 58 ° .