In a triangle ABC - the measure of - A is 40 - less then the measure of - B is 50 - less than that - C - Find the measure of - A

[ Let the measure of ∠ C be x^∘ #160;; measure of ∠ B=x 50^∘ #160;; measure of ∠ =x 90^∘ #160;; ...

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In a triangle...

Question

In a triangle $A B C,$ the measure of $∠ A$ is $40^{\circ}$ less then the measure of $∠ B$ is $50^{\circ}$ less than that $∠ C .$ Find the measure of $∠ A$

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Solution

$\begin{matrix} L e t t h e m e a s u r e o f ∠ C b e x^{\circ} m e a s u r e o f ∠ B = x - 50^{\circ} m e a s u r e o f ∠ = x - 90^{\circ} a / q 3 x - 140^{\circ} = 180^{\circ} x = 60 + \frac{140}{3} m e a s u r e o f ∠ A = \frac{140}{3} - 30 = \frac{50^{\circ}}{3} . \end{matrix}$

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In a triangle ABC, the measure of ∠A is 40∘ less then the measure of ∠B is 50∘ less than that ∠C. Find the measure of ∠A

Letthemeasureof∠Cbex∘measureof∠B=x−50∘measureof∠=x−90∘a/q3x−140∘=180∘x=60+1403measureof∠A=1403−30=50∘3.

In a triangle $A B C,$ the measure of $∠ A$ is $40^{\circ}$ less then the measure of $∠ B$ is $50^{\circ}$ less than that $∠ C .$ Find the measure of $∠ A$

$\begin{matrix} L e t t h e m e a s u r e o f ∠ C b e x^{\circ} m e a s u r e o f ∠ B = x - 50^{\circ} m e a s u r e o f ∠ = x - 90^{\circ} a / q 3 x - 140^{\circ} = 180^{\circ} x = 60 + \frac{140}{3} m e a s u r e o f ∠ A = \frac{140}{3} - 30 = \frac{50^{\circ}}{3} . \end{matrix}$