Question

this is the ∆PQR. find the ∆ QRP. find the value of X​

Answer 1

130°+y = 180° ( linear-pair )

y = 180° - 130°

y = 50°

x + 110° + 50° = 180° ( Angle Sum property )

160° + x = 180°

x = 180°-160°

x = 20°

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Answer 2

Answer

Find the value of x :-

In ∆PQR

P = 130°

y = ?

x =?

P = y [ Linear Pair ]

P + y = 180°

130° + y = 180°

y = 180° - 130°

y = 50°

P + Q + R = 180° [ angle of sum property ]

y + 110° + x = 180°

50° + 110° + x = 180°

x = 180° - 160°

x = 20°