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ABC is a right angles triangle with base BC and height AB. The hypotenuse AC is four times the length of the perpendicular drawn to it from opposite vertex. What is tan C equal to?
- a)2 – √3
- b)√3 – 1
- c)2 + √3
- d)√3 + 1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
ABC is a right angles triangle with base BC and height AB. The hypoten...
½ × AC × BD = ½ × AB × BC [Equating area]
AC × BD = AB × BC
4 BD × BD = AB × BC [∵ AC = 4 BD]
Dividing by BC2 both the sides
4 × BD/BC × BD/BD = AB/BC
In triangle BDC, sin C = BD/BC and In triangle ABC, tan C = AB/BC
4sin2C = tan C
4sin2C = sinC/cos C
4sinC cos C = 1
2sin C cos C = ½
Sin2C = ½ [∵ sin 2A = 2sinA cos A]
∴ 2C = 30° and C = 15°
tan C = tan 15°
To find tan 15°
Tan 2θ = 2tanθ/ (1 – tan2 θ)
Let tan 15° be x
Tan 30° = 2x/(1 – x2)
1/√3 = 2x/(1 – x2)
1 – x2 = 2√3x
x2 + 2√3x – 1 = 0
Using quadratic formula and ignoring the negative term
x = (-2√3 + 4)/2 = 2 – √3
4sin2C = tan C
4sin2C = sinC/cos C
4sinC cos C = 1
2sin C cos C = ½
Sin2C = ½ [∵ sin 2A = 2sinA cos A]
∴ 2C = 30° and C = 15°
tan C = tan 15°
To find tan 15°
Tan 2θ = 2tanθ/ (1 – tan2 θ)
Let tan 15° be x
Tan 30° = 2x/(1 – x2)
1/√3 = 2x/(1 – x2)
1 – x2 = 2√3x
x2 + 2√3x – 1 = 0
Using quadratic formula and ignoring the negative term
x = (-2√3 + 4)/2 = 2 – √3
Most Upvoted Answer
ABC is a right angles triangle with base BC and height AB. The hypoten...
Let the length of AB be x and the length of BC be y. Then, by the Pythagorean theorem, we have:
AC^2 = AB^2 + BC^2
(4x)^2 = x^2 + y^2
16x^2 = x^2 + y^2
15x^2 = y^2
We also know that AC is four times the length of the perpendicular drawn to it from opposite vertex. Let's call this perpendicular line segment DE, where D is on AC and E is on BC. Then:
DE = x
AC = 4x
AD = 3x
Now, we can use the tangent function to find tan C:
tan C = BC/AD
tan C = y/3x
But we know that y = 15x^2, so:
tan C = 15x^2/3x
tan C = 5x
We can find x by using the fact that AC = 4x:
16x^2 = x^2 + y^2
16x^2 = x^2 + (15x^2)
16x^2 = 16x^2
x = sqrt(16) = 4
Therefore, y = 15x^2 = 15(4^2) = 240.
Plugging in these values, we get:
tan C = 5x = 5(4) = 20
So, tan C is equal to 20.
AC^2 = AB^2 + BC^2
(4x)^2 = x^2 + y^2
16x^2 = x^2 + y^2
15x^2 = y^2
We also know that AC is four times the length of the perpendicular drawn to it from opposite vertex. Let's call this perpendicular line segment DE, where D is on AC and E is on BC. Then:
DE = x
AC = 4x
AD = 3x
Now, we can use the tangent function to find tan C:
tan C = BC/AD
tan C = y/3x
But we know that y = 15x^2, so:
tan C = 15x^2/3x
tan C = 5x
We can find x by using the fact that AC = 4x:
16x^2 = x^2 + y^2
16x^2 = x^2 + (15x^2)
16x^2 = 16x^2
x = sqrt(16) = 4
Therefore, y = 15x^2 = 15(4^2) = 240.
Plugging in these values, we get:
tan C = 5x = 5(4) = 20
So, tan C is equal to 20.
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