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3 years ago

In ∆ABC, m∠ACB = 90°, m∠A = 40°, and D ∈ AB such that CD is perpendicular to side AB. Find m∠DBC and m∠BCD.

Mathematics

1 answer:

podryga [215]3 years ago

8 0

Answer: ∠B = 50°

∠BCD = 40°

<u>Step-by-step explanation:</u>

ACB is a right triangle where ∠A = 40° and ∠C = 90°.

Use the Triangle Sum Theorem for ΔABC to find ∠B:

∠A + ∠B + ∠C = 180°

40° + ∠B + 90° = 180°

∠B + 130° = 180°

∠B = 50°

BCD is a right triangle where ∠B = 50° and ∠D = 90°.

Use the Triangle Sum Theorem for ΔBCD to find ∠C:

∠B + ∠C + ∠D = 180°

50° + ∠C + 90° = 180°

∠C + 140° = 180°

∠C = 40°

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The algebraic proof shows that the angles in an equilateral triangle must equal 60° each

<h3>Laws of cosines </h3>

From the question, we are to use the law of cosines to write an algebraic proof that shows that the angles in an equilateral triangle must equal 60°.

Given any triangle ABC, the measures of angles A, B, and C by the law of cosines are

cos A = (b^2 + c^2 - a^2)/2bc

cos B= (a^2 + c^2 - b^2)/2ac

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Now, given that the triangle is equilateral, with each of the side lengths equal to s

That is, a = b = c = s

Then, we can write that

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∴ A = cos⁻¹(0.5)

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Also

cos B = (s^2 + s^2 - s^2)/(2s×s)

cos B = (s^2 )/(2s^2)

cos B = 1/2

cos B = 0.5

∴ B = cos⁻¹(0.5)

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and

cos C = (s^2 + s^2 - s^2)/(2s×s)

cos C = (s^2 )/(2s^2)

cos C = 1/2

cos C = 0.5

∴ C = cos⁻¹(0.5)

C = 60°

Thus,

A = 60°, B = 60° and C = 60°

Hence, the algebraic proof above shows that the angles in an equilateral triangle must equal 60° each.

Learn more on The law of cosines here: brainly.com/question/2866347

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