Answer:
1) It is given that line AB is tangent to the circle at A.
∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)
Thus, the measure of ∠CAB is 90º.
1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:


Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹
Let <em>u</em> = (1, 0) and <em>v</em> = (0, 1). Then
<em>T</em> (<em>u</em>) = (2*1 - 3*0, 1 + 4, 5*0) = (2, 5, 0)
<em>T</em> (<em>v</em>) = (2*0 - 3*1, 0 + 4, 5*1) = (-3, 4, 5)
=> <em>T</em> (<em>u</em>) + <em>T</em> (<em>v</em>) = (-1, 9, 5)
but
<em>T</em> (<em>u</em> + <em>v</em>) = <em>T</em> (1, 1) = (2*1 - 3*1, 1 + 4, 5*1) = (-1, 5, 5)
=> <em>T</em> (<em>u</em> + <em>v</em>) ≠ <em>T</em> (<em>u</em>) + <em>T</em> (<em>v</em>)
which means <em>T</em> does not preserve addition, so it is not linear.
If secant of theta =
<span>
<span>
<span>
1.0416666667
</span>
</span>
</span>
then
theta = 16.26 Degrees
cotangent of theta = 3.4286