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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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yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

P(t) = 0.8t^{2} + 6t + 19000

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

P(t) = 0.8t^{2} + 6t + 19000

0.8t^{2} + 6t + 19000 = 34200

0.8t^{2} + 6t - 15200 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

0.8t^{2} + 6t - 15200 = 0

So a = 0.8, b = 6, c = -15200

Then

\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356

t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14

t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

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