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Sergeu [11.5K]
3 years ago
13

Solve tan 10 - tan 50 +tan 70 with trigonometry.

Mathematics
1 answer:
Jobisdone [24]3 years ago
7 0
Below is the solution, I hope it helps.

  <span>i) tan(70) - tan(50) = tan(60 + 10) - tan(60 - 10) 

= {tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)} 

ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to: 

= 8*tan(10)/{1 - 3*tan²(10)} 

iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10) 

= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} 

= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} 

= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)} 

= 3*tan(30) = 3*(1/√3) = √3 [Proved] 

[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)}, 
{3*tan(10) - tan³(10)}/{1 - 3*tan²(10)} = tan(3*10) = tan(30)]</span>
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Marianna [84]

Answer:

  B. {16, 19, 20}

Step-by-step explanation:

The <em>triangle inequality</em> requires for any sides a, b, c you must have ...

  a + b > c

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The net result of those requirements are ...

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If we look at the offered side length choices, we see ...

  A: 8+11 = 19 . . . not > 19; not a triangle

  B: 16+19 = 35 > 20; could be a triangle

  C: 3+4 = 7 . . . not > 8; not a triangle

  D: 5+5 = 10 . . . not > 11; not a triangle

The side lengths {16, 19, 20} could represent the sides of a triangle.

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<em>Additional comment</em>

The version of triangle inequality shown above ensures that a triangle will have non-zero area.

The alternative version of the triangle inequality uses ≥ instead of >. Triangles where a+b=c will look like a line segment--they will have zero area. Many authors disallow this case. (If it were allowed, then {8, 11, 19} would also be a "triangle.")

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