(1) In △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively. If a = 8, ∠ B = 60 ° and ∠ C = 75 °, find ∠ A and B; (2) in △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively. A = 4, B = 5 and C = 61 are known

(1) In △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively. If a = 8, ∠ B = 60 ° and ∠ C = 75 °, find ∠ A and B; (2) in △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively. A = 4, B = 5 and C = 61 are known

(1) ∵∵ B = 60 °, ∵ C = 75 °, ∵ a = 45 °; b = asinbsina = 8 · sin60 ° sin45 ° = 46; (2) ①∵ a = 4, B = 5, C = 61, ∵ COSC = A2 + b2-c22ab = 42 + 52 - (61) 22 · 4 · 5 = - 12, ∵ 0 ° & lt; C & lt; 180 °, ∵ C = 120 ° can be obtained from the sine theorem; (2) sinc = 32, ∵ s △ ABC = 12absinc = 12 × 4 × 5 × 32 = 53

In the space quadrilateral PABC, if PA, Pb and PC are perpendicular, it is proved that the triangle ABC must be an acute angle
Be specific

The dihedral angles defined by PA, Pb and PC are all straight dihedral angles PA.CE Let AE ≤ ad (E and D are close to a) make ef ⊥ AP, f ∈ AB on PAB. Look at ⊥ CEF and ⊥ CAE, CF & sup2; = CE & sup2; + EF & sup2; = Ca & sup2; + AF & sup2; - 2ca × AF × cos ∠ bac

Let p be any point in the equilateral triangle ABC, and prove PA < Pb + PC

∵PB+PC>BC
And P is a point in the triangle,
∴PA

A flea jumps in the first quadrant, namely x-axis and y-axis. In the first second, he jumps from the origin to (0,1), and then jumps in the direction shown by the arrow in the figure [i.e
（0,0）→（0,1）→（1,1）→（1,0）→… ], and jump one unit per second, then the coordinates of the flea's position in the 35th second are____

Comrade, four seconds is a cycle, right
35 / 4 = 8,3
The coordinates of the last point in each cycle are the coordinates of the first point plus two
8*2=16
The third second (the fourth point) moves one space to the right relative to the first point
So 35 seconds to the point (17,0)
Children's shoes don't know how to ask

A flea jumps in the first quadrant, that is, the x-axis and y-axis. In the first second, he jumps from the origin to (0,1), and then jumps in the direction shown by the arrow in the figure
[i.e. (0,0) → (0,1) → (1,1) → (1,0) → ], and jump one unit per second, then the coordinates of the flea's position in the 35th second are__________

35 / 4 = 8,3
The coordinates of the last point in each cycle are the coordinates of the first point plus two
8*2=16
The third second (the fourth point) moves one space to the right relative to the first point
So 35 seconds to the point (17,0)

A flea jumps in the first quadrant, x-axis and y-axis. In the first second, it jumps from the origin to (0,1), and then jumps in the direction indicated by the arrow in the figure [i.e. (0,0) → (0,1) → (1,1) → (1,0) →...] ], and jump one unit per second, then the coordinates of the flea's position in the 2014 second are (& nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;) and the nth second are (& nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;)

It takes two seconds to reach point (1,1)
It takes 2 + 4 seconds to reach (2,2)
It takes 2 + 4 + 6 seconds to reach (3,3)
It takes 2 + 4 + 6 +... + 2n seconds to reach (n, n) = n (n + 1) seconds

Log reduction! Log2 (3) [is the index of (& frac12;)] is reduced by one third,

y=(1/2)^(log2(3)
Then take 1 / 2 logarithm on both sides at the same time
log(1/2)(y)=log(2)(3)=-log(1/2)(3)=log(1/2)(1/3)
y=1/3

Log (a + 3) ^ (a ^ 2-A) = 1 to find the value of a

Domain: a + 3 > 0 and a + 31, a ^ 2-A > 0
(a+3)=(a^2-a)
a^2-2a-3=0
A = - 1 or 3

If two thirds of log a is greater than 1, the value range of a is?

A is more than two thirds

How to simplify log (2x) + log (3Y) - log (2Z)? Log3 (x ^ 2-2x-6) = 2 log (3x + 6) = 1 + log (x) is solved
How to simplify log (2x) + log (3Y) - log (2Z)? Log3 (x ^ 2-2x-6) = 2 log (3x + 6) = 1 + log (x) is solved,

log(2x)+log(3y)-log(2z)=log(3xy/z)
log3(x^2-2x-6)=2
log3(x^2-2x-6)=log3(9)
x^2-2x-6=9
X = - 3 or x = 5
log(3x+6)=1+log(x)=log(10x)
3x+6=10x
x=6/7