A ray $O C$ stands on a line $A O B$ . Find $x$ if $∠ A O C = x + 20^{o}$ and $∠ B O C = 3 x - 8^{o}$ .

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Solution

$\begin{matrix} F r o m t h e f i g u r e w e g e t ∠ A O C + ∠ B O C = 180^{\circ} \Rightarrow x + 20^{\circ} + 3 x - 8^{\circ} = 180^{\circ} \Rightarrow 4 x = 180^{\circ} - 12^{\circ} \Rightarrow x = \frac{168^{\circ}}{4} ∴ x = 42^{\circ} h e n c e, t h e v a l u e o f x i s 42^{\circ} . \end{matrix}$

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A ray OC stands on a line AOB. Find x if ∠AOC=x+20o and ∠BOC=3x−8o.

Fromthefigureweget∠AOC+∠BOC=180∘⇒x+20∘+3x−8∘=180∘⇒4x=180∘−12∘⇒x=168∘4∴x=42∘hence,thevalueofxis42∘.

A ray $O C$ stands on a line $A O B$ . Find $x$ if $∠ A O C = x + 20^{o}$ and $∠ B O C = 3 x - 8^{o}$ .