Question

If ∆ABC=∆PQR and angle ABC=60° then find angle PQR​

Answer 1

Step-by-step explanation:

$\huge \boxed{\underbrace{\mathfrak {\underline {\fcolorbox {black}{pink}{Answer}}}}}$

${\underline{\underline { <strong>Given</strong><strong> </strong> }}}$

${ \triangle {ABC}=\triangle {PQR} }$

and.${ \angle {ABC}=60° }$

.

${\underline{\underline { <strong>find</strong><strong> </strong> }}}$

${ \angle {PQR=????} }$

$\huge\boxed {\pink {\mathbb{\underline {<strong>Answer</strong><strong> </strong>}}}}$

${ here....\triangle{ ABC }=\triangle {PQR} }$

$\implies { so....\triangle {ABC}~\triangle {PQR} }$

$\implies { \angle {ABC}=\angle {PQR} }$

$\implies {but....\angle {ABC}=60° }$

$\implies { then...\angle {PQR}=60° }$

Answer 2

Step-by-step explanation:

Given$</p><p>{ \triangle {ABC}=\triangle {PQR} }△ABC=△PQR</p><p>and.{ \angle {ABC}=60° }∠ABC=60°</p><p>.</p><p>{\underline{\underline { find }}}find</p><p>{ \angle {PQR=????} }∠PQR=????</p><p>\huge\boxed {\pink {\mathbb{\underline {Answer }}}}Answer</p><p>{ here....\triangle{ ABC }=\triangle {PQR} }here....△ABC=△PQR</p><p>\implies { so....\triangle {ABC}~\triangle {PQR} }⟹so....△ABC △PQR</p><p>\implies { \angle {ABC}=\angle {PQR} }⟹∠ABC=∠PQR</p><p>\implies {but....\angle {ABC}=60° }⟹but....∠ABC=60°</p><p>\implies { then...\angle {PQR}=60° }⟹then...∠PQR=60°$