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keving4three
05.06.2020 •
Mathematics
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
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Ответ:
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
Ответ:
-10
Step-by-step explanation:
I used a calculator lol