Question

if ∆PQR ≈ ∆ABC , then angles P =50° , angle


Q= 70° Find angle A and B of ∆ ABC​

Answer 1

Answer:

In △PQR using angle sum property of triangle.

∠P+∠Q+∠R=180

∘

∠P+50

∘

+70

∘

=180

∘

⇒∠P=180

∘

−120

∘

=60

∘

Now △PQR∼△XYZ

As corresponding angles of similar triangles are equal.

⇒∠X=∠P=60

∘

and ∠Y=∠Q=50

∘

⇒∠X+∠Y=60

∘

+50

∘

=110

∘

Answer 2

Answer:

In △PQR using angle sum property of triangle.

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZ

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘ ⇒∠X+∠Y=60

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘ ⇒∠X+∠Y=60 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘ ⇒∠X+∠Y=60 ∘ +50

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘ ⇒∠X+∠Y=60 ∘ +50 ∘

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘ ⇒∠X+∠Y=60 ∘ +50 ∘ =110

In △PQR using angle sum property of triangle.∠P+∠Q+∠R=180 ∘ ∠P+50 ∘ +70 ∘ =180 ∘ ⇒∠P=180 ∘ −120 ∘ =60 ∘ Now △PQR∼△XYZAs corresponding angles of similar triangles are equal.⇒∠X=∠P=60 ∘ and ∠Y=∠Q=50 ∘ ⇒∠X+∠Y=60 ∘ +50 ∘ =110 ∘

I am not sure about this answer...