Question

In △ABC, ∠B=20° and ∠C=40°. The angle bisector at A intersects side BC at point D. What is BC - AB if AD = 24

Answer 1

Answer:

hi

Explanation:

In triangle ABD AD/sin 20 =BD/sin 60 ; BD= 1*sin 60/sin 20 =0.8660/0.3420 =2.53

In triangle ADC CD/sin 60 =AD/sin 40 ; CD= 1*sin 60/sin 40 = 0.8660/0.6428 =1.35

BD+DC= 2.53+1.35=3.88

Now in Triangle ABC

3.88/sin 120 =AB/sin 40 ; AB=3.88*sin 40/cos 3 0 [∵sin[90+30) =cos 30]

=3.88*0.6428/0.8660 =2.88

BC-AB =3.88–2.88 = 1

Answer: The difference between BC and AB is = 1 unit

Answer 2

Explanation:

In △ABC, ∠B=20° and ∠C=40°. The angle bisector at A intersects side BC at point D. What is the difference between BC and AB if AD = 1?

I don’t see a short cut. So Law of Sines will solve triangles ABD and ACD.

In triangle ABD AD/sin 20 =BD/sin 60 ; BD= 1*sin 60/sin 20 =0.8660/0.3420 =2.53

In triangle ADC CD/sin 60 =AD/sin 40 ; CD= 1*sin 60/sin 40 = 0.8660/0.6428 =1.35

BD+DC= 2.53+1.35=3.88

Now in Triangle ABC

3.88/sin 120 =AB/sin 40 ; AB=3.88*sin 40/cos 3 0 [∵sin[90+30) =cos 30]

=3.88*0.6428/0.8660 =2.88

BC-AB =3.88–2.88 = 1

Answer: The difference between BC and AB is = 1 unit