QUESTION IMAGE
Question
- t 5x + 10 58° s 11x + 2 u a 11) c 46° -1 + 8x d 18x + 5 b w 12) c 10x - 10 a 12x - 4 q 30° b 13) c 15x + 5 120° a d 22x + 4 b 14) d 9x - 2 b 20x + 5 v 40° c find the measure of the angle indicated.
Step1: Recall exterior - angle property of a triangle
The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
Step2: Solve problem 10
In triangle \(TSU\), the exterior angle \(\angle TAU=11x + 2\), and the non - adjacent interior angles are \(5x+10\) and \(58^{\circ}\). So, \(11x + 2=(5x + 10)+58\).
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The angle \(\angle T = 5x+10=5\times11 + 10=65^{\circ}\).
Step3: Solve problem 11
In triangle \(BCD\), the exterior angle \(\angle CBW = 18x+5\), and the non - adjacent interior angles are \(46^{\circ}\) and \(- 1+8x\). So, \(18x+5=46+( - 1+8x)\).
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The angle \(\angle D=-1 + 8x=-1+8\times4 = 31^{\circ}\).
Step4: Solve problem 12
The exterior angle \(\angle CAQ=12x - 4\), and the non - adjacent interior angles are \(10x-10\) and \(30^{\circ}\). So, \(12x - 4=(10x-10)+30\).
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The angle \(\angle CAB=12x-4=12\times12-4 = 140^{\circ}\).
Step5: Solve problem 13
The exterior angle \(\angle ADB = 120^{\circ}\), and the non - adjacent interior angles are \(15x+5\) and \(22x + 4\). So, \(120=(15x+5)+(22x + 4)\).
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The angle \(\angle C=15x+5=15\times3+5 = 50^{\circ}\).
Step6: Solve problem 14
The exterior angle \(\angle DBV=20x + 5\), and the non - adjacent interior angles are \(9x-2\) and \(40^{\circ}\). So, \(20x + 5=(9x-2)+40\).
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The angle \(\angle CDB=9x-2=9\times3-2 = 25^{\circ}\).
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Problem 10: \(\angle T = 65^{\circ}\)
Problem 11: \(\angle D = 31^{\circ}\)
Problem 12: \(\angle CAB=140^{\circ}\)
Problem 13: \(\angle C = 50^{\circ}\)
Problem 14: \(\angle CDB=25^{\circ}\)