Question-222753

Question Number 222753 by mr W last updated on 06/Jul/25

Answered by A5T last updated on 07/Jul/25

Let θ be the base angle of any of the isosceles △ 180−2θ+180−(2(180−θ−?))+90=180 ⇒2?=90⇒?=45°

$$\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{base}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}\:\mathrm{isosceles}\:\bigtriangleup \\ $$$$\mathrm{180}−\mathrm{2}\theta+\mathrm{180}−\left(\mathrm{2}\left(\mathrm{180}−\theta−?\right)\right)+\mathrm{90}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{2}?=\mathrm{90}\Rightarrow?=\mathrm{45}° \\ $$

Commented by mr W last updated on 07/Jul/25

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Answered by fantastic last updated on 07/Jul/25

Commented by fantastic last updated on 07/Jul/25

λ=180^0 −2θ ψ=180^0 −(90^0 +λ)=90^0 −180^0 +2θ=2θ−90^0 α=((180^0 +90^0 −2θ)/2)=((270^0 −2θ)/2)=135^0 −θ β=180^0 −(α+θ)=180^0 −(135^0 −θ+θ)=180^0 −135^0 =45^0 ✓

$$\lambda=\mathrm{180}^{\mathrm{0}} −\mathrm{2}\theta \\ $$$$\psi=\mathrm{180}^{\mathrm{0}} −\left(\mathrm{90}^{\mathrm{0}} +\lambda\right)=\mathrm{90}^{\mathrm{0}} −\mathrm{180}^{\mathrm{0}} +\mathrm{2}\theta=\mathrm{2}\theta−\mathrm{90}^{\mathrm{0}} \\ $$$$\alpha=\frac{\mathrm{180}^{\mathrm{0}} +\mathrm{90}^{\mathrm{0}} −\mathrm{2}\theta}{\mathrm{2}}=\frac{\mathrm{270}^{\mathrm{0}} −\mathrm{2}\theta}{\mathrm{2}}=\mathrm{135}^{\mathrm{0}} −\theta \\ $$$$\beta=\mathrm{180}^{\mathrm{0}} −\left(\alpha+\theta\right)=\mathrm{180}^{\mathrm{0}} −\left(\mathrm{135}^{\mathrm{0}} −\theta+\theta\right)=\mathrm{180}^{\mathrm{0}} −\mathrm{135}^{\mathrm{0}} =\mathrm{45}^{\mathrm{0}} \checkmark \\ $$

Commented by mr W last updated on 07/Jul/25

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Answered by mr W last updated on 07/Jul/25

Commented by mr W last updated on 07/Jul/25

$$?={x}=\alpha+\beta=\frac{\mathrm{2}\alpha+\mathrm{2}\beta}{\mathrm{2}}=\frac{\mathrm{90}°}{\mathrm{2}}=\mathrm{45}° \\ $$

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