Delta XYZ is right-angled at Y. If m angle X | vedclass

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(A) In a triangle $\Delta XYZ$,the sum of all interior angles is $180^{\circ}$.Given that $\angle Y = 90^{\circ}$ and $\angle X = 60^{\circ}$.Therefor

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}$

Vedclass · Answer

(A) In a triangle $\Delta XYZ$,the sum of all interior angles is $180^{\circ}$.Given that $\angle Y = 90^{\circ}$ and $\angle X = 60^{\circ}$.Therefore,$\angle Z = 180^{\circ} - (90^{\circ} + 60^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.Now,we need to find the value of $\sec Z + \frac{2}{\sqrt{3}}$.Since $\angle Z = 30^{\circ}$,$\sec Z = \sec 30^{\circ} = \frac{1}{\cos 30^{\circ}} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$.Substituting this value into the expression:$\sec Z + \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

$\Delta XYZ$ is right-angled at $Y$. If $m \angle X = 60^{\circ}$,then find the value of $\left(\sec Z + \frac{2}{\sqrt{3}}\right)$.

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