In ∆PQR angleP =2angleQ and 2angleR=3angleQ find all angles of triangle
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given, P=2Q, 2R=3Q
let angle Q be "x"
P=2x,Q=x ,and R=(3/2)x
sum of all angles of a triangle is 180°
so 2x+x+(3÷2)x =180°
from here we get that x=40°
Q=40,P=80and R=60
let angle Q be "x"
P=2x,Q=x ,and R=(3/2)x
sum of all angles of a triangle is 180°
so 2x+x+(3÷2)x =180°
from here we get that x=40°
Q=40,P=80and R=60
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1
Answer:
Step-by-step explanation:
given, P=2Q, 2R=3Q
let angle Q be "x"
P=2x,Q=x ,and R=(3/2)x
sum of all angles of a triangle is 180°
so 2x+x+(3÷2)x =180°
from here we get that x=40°
Q=40,P=80and R=60
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