I just looked through the books.

1: AB = vt = 15 nm /h* 1.75h=26.25 nm.

? The outer angle of the triangle is ∠ C+∠c+∠a = 52°.

? ∫∠A = 26

? ∴∠C=26 =∠A

? BC = ∴BC=AB=26.25 nautical miles

2: It is an isosceles triangle.

? * AD is parallel to BC.

? ∴∠DAC=∠C,∠EAD=∠B

? ∫AD bisection △ABC external angle ∠EAC

∴∠EAD=∠DAC=∠B=∠C

? ∴AB=AC

You can ask me later if you don't understand. ?

Subtitle: 1,

As shown in the figure, at 8: 00 a.m., Ship A set out from A, sailed due north at the speed of 15 knots, and arrived at B at 9: 40. From A, lighthouse C is 26 in the northwest, and from B, lighthouse C is 52 in the northwest, so the distance from B to lighthouse C is ().

2、

As shown in the figure, given that AD divides ∠EAC and ABC equally, then △ABC must be?

Triangle.