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Bulb root mathematics
1. solution: in isosceles △ABC,

AB? =? AC? =? 6,

Through point a as AD⊥BC in d,

AD? =? h,

BD? =? DC? =? x,

Judging from the title:

xh= 12,

From Pythagorean Theorem: X? +? h^? =? 36,

Put xh? =? 24 into x? +? h^? =? 36 Must:

x^? +? h^? =? 3xh/2,

∴2h^? -? 3xh? -? 2x^? =? 0,

Divide both sides by x:

2(h/x)^? -? 3h/x? -? 2? =? 0,

Take (h/x) as a whole, and get it by finding the root formula:

h/x? =? 2 (I will omit the negation)

∴ The tangent of the base angle is 2,

What they did was wrong! )

2. Make a symmetrical point P of point E about AC. Obviously, p is on AD.

Connect PB to AC to f,

EF at this time? +? BF? At the very least,

Number three, number three,

3. In Rt△ABC, ∠ c = 90,

Let the opposite sides of angles a, b and c be a, b and c respectively.

∴sinA? =? Air conditioning?

Tana? =? a/b

∵ Sina Yi? +? Tana 1 =5

∴c/a? +? b/a? =? 5,

∴b? +? c? =? 5a,

From Pythagorean Theorem:

a^? +? b^? =? c^,

Play b? +? c? =? Replace a with 5a? +? b^? =? Merge and simplify:

13(b/c)^? +? b/c? -? 12? =? 0,

Considering b/c as a whole, it is obtained from the bulbous formula:

b/c? =? 12/ 13,

∴cosA? =? 12/ 13,

4. Solution: ∵ The zeroth power of any non-zero real number is 1.

∴(√3- 1) is 1.

The answer is detailed enough!