The solution idea of 1 is to verify the side lengths of AC and AB through the sine values of angle B and angle C.

Solution: make BC's vertical line pass through point A: AD, sinB=AD/AB, sinC=AD/AC,

Because the angle b > Angle c, so sinB & gtSinC, AD/AB & gt；; AD/AC，

Get: ab

Maybe the landlord should post the picture of the second question to be complete and clear. Let me guess for a moment, the picture should be:

Two triangles: ABC, PBC, PB are the inner bisector of angle ABC, and PC is the outer bisector of angle ACB.

If the speculation is correct, then the idea to solve the second problem is to use a formula to solve the problem by using the relationship that the sum of the angles in the triangle is 180, the relationship between the angle ABC and the angle PBC, and the relationship between the angle ACB and the angle PCB.

Solution: angle ABC = 2° angle PBC (according to the concept of internal angle bisector)

Angle PCB= angle PCA+ angle ACB,

And the angle PCA= 1/2( 180- angle ACB) (according to the concept of outer angle and bisector of outer angle).

Angle ABC+ Angle ACB= 180-64 (according to the concept of triangle interior angle sum)

Angle P= 180- Angle PBC- Angle PCB (according to the concept of the sum of the internal angles of a triangle)

The solution can be obtained by combining the above formulas. I believe the landlord will understand after a little thinking.