Question

if one angle of triangle is 48 ° and the other two angles are in the ratio 6 :5then the largest angle is​

Answer 1

$\underline{ \huge \frak{Given : : }}$

• One angle of triangle is 48°
• Ratio of other angles of triangle = 6 : 5

$\underline{ \huge \frak{To \: }{ \frak{find : : }}}$

• Largest angle of the triangle.

$\underline{ \huge \frak{SoluTion : :}}$

$\implies6x + 5x + 48 \degree= 180\degree \\ \implies \: \: \: \: \: \: \: \: 11x + 48\degree = 180\degree \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 11x = 180\degree - 48\degree \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{180\degree - 48\degree}{11} \\ \boxed{\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{132\degree}{11} =12\degree }$

Hence,

$\sf \: {1}^{st} \: \: angle = 48\degree \\ \sf {2}^{nd} \: \: angle = 72\degree \\ \sf {3}^{rd} \: \: angle = 60\degree$

Thus,

$\red{ \bf\bullet \: \: 72\degree \: \: is \: \: the \: \: largest \: \: angle \: \: of }\\ \red{ \bf \: \: \: \: the \: \: triangle.}$

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