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4). Triangle ABC. Angle B is 9 times smaller than Angle A. Find Angle B and Angle A.
Вопрос:
4). Triangle ABC. Angle B is 9 times smaller than Angle A. Find Angle B and Angle A.
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Solution:
- Let angle A be represented by 'x'.
- According to the problem, angle B is 9 times smaller than angle A, so angle B = x / 9.
- The triangle ABC is a right triangle with the right angle at C (indicated by the square symbol). Therefore, angle C = 90 degrees.
- The sum of angles in a triangle is 180 degrees.
- So, angle A + angle B + angle C = 180 degrees.
- Substitute the expressions for angles A and B, and the value of angle C into the equation: x + (x / 9) + 90 = 180.
- Subtract 90 from both sides: x + (x / 9) = 180 - 90.
- x + (x / 9) = 90.
- To add x and x/9, find a common denominator, which is 9: (9x / 9) + (x / 9) = 90.
- Combine the terms: (9x + x) / 9 = 90.
- 10x / 9 = 90.
- To solve for x, multiply both sides by 9: 10x = 90 * 9.
- 10x = 810.
- Divide both sides by 10: x = 810 / 10.
- x = 81. So, angle A = 81 degrees.
- Now, find angle B: angle B = x / 9 = 81 / 9 = 9 degrees.
- Let's check if the sum of angles is 180: Angle A + Angle B + Angle C = 81° + 9° + 90° = 180°. The sum is correct.
Ответ: ∠B = 9°, ∠A = 81°
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