Question

angles of triangle are (x-40), (x-20) and 1/2 (x-10) find the value of x​

Answer 1

Answer:

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Answer 2

$\huge\underline\mathfrak\green{Solution:}$

Since (x-40°), (x-20°) and $\bf\left(\frac{x}{2}-10\right)$ are the angles of a ∆, therefore

$\bf{(x-40°)+(x-20°)+}$$\bf\left(\frac{x}{2}-10°\right)=180°$

$\implies$$\bf{x-40°+x-20°+}$$\bf\left(\frac{x}{2}\right)-10°=180°$

$\implies$$\bf{2x+}$$\bf\left(\frac{x}{2}\right)-70°=180°$

$\implies$$\bf\frac{5}{2}x=180°+70°$

$\implies$$\bf\frac{5}{2}x=250°$

$\implies$$\bf x=250°\times\frac{2}{5}$

$\hookrightarrow$$\bf{x=50°\times2}$

$\hookrightarrow$$\bf\underline{x=100°}$

$\underline\mathfrak\pink{Hence,\:the\:angles\:of\:the\:triangles\:are:}$

$\bf{(x-40°)=(100-40)^°=60°}$

$\bf{(x-20°)=(100-20)^°=80°}$

$\bf\left(\frac{x}{2}-10°\right)=\left(\frac{100}{2}-10°\right)=(50°-10)^°=40°$