HELP PLEASE!! Given: Isosceles triangle ABC with vertex angle A, equal sides AB and AC, and an angle bisector, BD.

side = 1
angle = 36

BC has a length of 1. To the nearest tenth of a unit, find the lengths of the missing sides.

BD = __ units

AD = __ units

AB = __ units

AC = __ units

BD = 1 unit

AD = 1 unit

AB = 1.6 units

AC = 1.6 units

Step-by-step explanation:

In the picture attached, the triangle ABC is shown.

Given that triangle ABC is isosceles, then ∠B = ∠C

∠A + ∠B + ∠C = 180°

36° + 2∠B = 180°

∠B = (180° - 36°)/2

∠B = ∠C  = 72°

From Law of Sines:

sin(∠A)/BC = sin(∠B)/AC = sin(∠C)/AB

(Remember that BC is 1 unit long)

AB = AC = sin(72°)/sin(36°) = 1.6

In triangle ABD, ∠B = 72°/2 = 36°, then:

∠A + ∠B + ∠D = 180°

36° + 36° + ∠D = 180°

∠D =  180° - 36° - 36° = 108°

From Law of Sines:

sin(∠A)/BD = sin(∠B)/AD = sin(∠D)/AB

(now ∠A = ∠B)

BD = AD = sin(∠A)*AB/sin(∠D)

BD = AD = sin(36°)*1.6/sin(108°) = 1

$6.42

step-by-step explanation:

a: inscribed angles

step-by-step explanation:

i reviewed this question with a peer and teacher, seems to be correct, me later