The measure of the angles of a pentagon are $(x - 5)^{0}, (x - 6)^{0}, (2 x - 7)^{0}, x^{0}$ and $(2 x - 2)^{0} .$ Find value of $x$ .

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Solution

We know that the sum of all angles of the pentagon

$= 540^{\circ}$

Therefore,

${(x - 5)}^{\circ} + {(x - 6)}^{\circ} + {(2 x - 7)}^{\circ} + x^{\circ} + {(2 x - 2)}^{\circ} = 540^{\circ}$

$7 x^{\circ} - 20^{\circ} = 540^{\circ}$

$7 x^{\circ} = 560^{\circ}$

$x^{\circ} = 80^{\circ}$

Hence, this is the answer.

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The measure of the angles of a pentagon are (x−5)0,(x−6)0,(2x−7)0,x0 and (2x−2)0. Find value of x.

We know that the sum of all angles of the pentagon =540∘ Therefore, (x−5)∘+(x−6)∘+(2x−7)∘+x∘+(2x−2)∘=540∘ 7x∘−20∘=540∘ 7x∘=560∘ x∘=80∘ Hence, this is the answer.

The measure of the angles of a pentagon are $(x - 5)^{0}, (x - 6)^{0}, (2 x - 7)^{0}, x^{0}$ and $(2 x - 2)^{0} .$ Find value of $x$ .