Given : tan 235 = 2 tan 20 + tan 215
To Find : prove that
Solution:
tan 235 = 2 tan 20 + tan 215
Tan x = Tan (180 + x)
tan 235 = tan ( 180 + 55) = tan55
tan 215 = tan (180 + 35) = tan 35
=> tan 55 = 2tan 20 + tan 35
55 = 20 + 35
=> 20 = 55 - 35
taking Tan both sides
=> Tan 20 = Tan ( 55 - 35)
=> Tan 20 = (Tan55 - Tan35) /(1 + Tan55 . Tan35)
Tan35 = Cot55 = 1/tan55 => Tan55 . Tan35 =1
=> Tan 20 = (Tan 55 - Tan 35) /(1 + 1)
=> Tan 20 = (Tan 55 - Tan 35) /2
=> 2 Tan 20 = Tan 55 - Tan 35
=> 2 Tan 20 + Tan 35 = Tan 55
=> tan 55 = 2tan 20 + tan 35
=> tan 235 = 2tan 20 + tan 215
QED
Hence Proved
<span><span><u>Answer</u>
This explanation shows that the equation is solvable and there is not where in the steps we get a zero on the denominator.
</span><span><u>Explanation</u>
</span><span>The equation given in the statement is;
4/5+3/x=1/2
This equation can be solved as follows:
4/5+3/x=1/2
3/x=1/2-4/5=(5-8)/10
3/x= -3/10
Taking the reciprocal on both side of the equation;
x/3=-10/3
Multiplying by 3 both sides we get;
</span><span>x= -10
</span></span>
C. $360
$224x4=896 (total profit)
$896 (total) - $536 (first month profit) = $360 (second month profit)
(9,40,41) is a Pythagorean Triple, farther down the list than teachers usually venture.
Answer: D. 41 cm
There's a subset of Pythagorean Triples where the long leg is one less than the hypotenuse,
a^2+b^2 = (b+1)^2
a^2 + b^2 = b^2 + 2b +1
a^2=2b+1
So we get one for every odd number, since the square of an odd number is odd and the square of an even number is even.
b = (a^2 - 1)/2
a=3, b=(3^2-1)/2=4, c=b+1=5
a=5, b=(5^2-1)/2 =12, c = 13
a=7, b=24, c=25
a=9, b=40, c=41
a=11, b=60, c=61
a=13, b=84, c=85
It's good to be able to recognize Pythagorean Triples when we see them.
Otherwise we'd have to work the calculator:
√(9² + 40²) = √1681 = 41
