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ABC is a right angled triangle with base BC and height AB. The hypotenuse AC is four times the length of the perpendicular drawn to it from the opposite vertex. What is tan C equal to?
Let AB = a and BC = b.
Now, 2 cases may be possible in this question
Case I: a>b
Case II: a In triangle ABC, Area = ½ AB*BC = ½ * OB*AC => ½ ab = ½ x*4x => ab = 4x2 => 2ab = 8x2 Applying Pythagoras theorem in this triangle, a2 + b2 = (4x)2 = 16x2 Now, (a + b)2 = a2 + b2 + 2ab => (a + b)2 = 16x2 + 4x2 = 26x2 Thus, (a + b) = 2√6 x Similarly, (a - b)2 = a2 + b2 - 2ab Thus, (a – b) = 2√2 x and -2√2 x (Considering Case I & II mentioned above) Now, Case I: (a + b) + (a - b) = 2√6 x + 2√2 x & (a + b) - (a - b) = 2√6 x - 2√2 x => a = (√6 + √2) x & b = (√6 - √2) x Case II: (a + b) + (a - b) = 2√6 x - 2√2 x & (a + b) - (a - b) = 2√6 x + 2√2 x => a = (√6 - √2) x & b = (√6 + √2) x Now, Tan C = a/b So, Tan c = (√6 - √2) x/ (√6 - √2) x (Case I) On rationalizing, Tan C = 2 + √3 For Case II, Tan C = 2 - √3 so Tan C = 2+√3 or Tan C = 2 - √3
In triangle ABC, Area = ½ AB*BC = ½ * OB*AC => ½ ab = ½ x*4x => ab = 4x2 => 2ab = 8x2
Applying Pythagoras theorem in this triangle,
a2 + b2 = (4x)2 = 16x2
Now, (a + b)2 = a2 + b2 + 2ab => (a + b)2 = 16x2 + 4x2 = 26x2
Thus, (a + b) = 2√6 x Similarly, (a - b)2 = a2 + b2 - 2ab Thus, (a – b) = 2√2 x and -2√2 x (Considering Case I & II mentioned above)
Now, Case I: (a + b) + (a - b) = 2√6 x + 2√2 x & (a + b) - (a - b) = 2√6 x - 2√2 x => a = (√6 + √2) x & b = (√6 - √2) x
Case II: (a + b) + (a - b) = 2√6 x - 2√2 x & (a + b) - (a - b) = 2√6 x + 2√2 x => a = (√6 - √2) x & b = (√6 + √2) x
Now, Tan C = a/b So, Tan c = (√6 - √2) x/ (√6 - √2) x (Case I)
On rationalizing, Tan C = 2 + √3 For Case II, Tan C = 2 - √3
so Tan C = 2+√3 or Tan C = 2 - √3
By: Munesh Kumari ProfileResourcesReport error
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