Question

In the figure find the angle​

Answer 1

$\huge \bf{ \blue{Hey \: there!!}}$

$\huge{\purple{\underline{\red{\underline{\pink{\pmb{\mathfrak{ given}}}}}}}}$

PQR is a Triangle.

∠R = 140⁰

$\huge{\purple{\underline{\red{\underline{\pink{\pmb{\mathfrak{ to \: find}}}}}}}}$

$∠x .∠y.∠p$

$\huge{\purple{\underline{\red{\underline{\pink{\pmb{\mathfrak{ solution}}}}}}}}$

$∠y = 180 ^{0} - {140}^{0} (linear \: pair) \\ \purple{ \boxed{∠y = {40}^{0}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$∠y = ∠x \: (opposite \: sides \: of \: isoceles \: triangle \: are \: equal)$

$\purple{ \boxed{∠x = {40}^{0} }}$

$\tt \red{{by \: angle \: sum \: property \: of \: a \: triangle}}$

$∠y + ∠x + ∠p= {180}^{0}$

${40}^{0} + {40}^{0} + ∠p = {180}^{0}$

$80 ^{0} + ∠p = {180}^{0}$

$∠p = {180}^{0} - {80}^{0}$

$\purple{ \boxed{∠p = 100}}$

• I Hope it's Helpful My Friend.