Question

Given: ∆ABC, m∠C = 90° m∠BAC = 2m∠ABC BC = 24, AL −∠ bisector Find: AL

Answer 1

M∠BCA=90, then m∠BAC +∠ABC=90
m∠BAC = 2m∠ABC, then  2m∠ABC + m∠ABC=90, m∠ABC=30, and m∠CAB=60

ΔBAC is a 30-60-90 right triangle, in which the ratio of the side lengths is 1:√3:2
AC:BC=1:√3, AC=BC/√3
BC=24, AC=24/√3=8√3, (you don't need AB, but AB=2AC=16√3)

AL bisects angle A =>m∠LAC=30
ΔALC is a 30-60-90 right triangle, in which the ratio of the side lengths is 1:√3:2
AC:AL=√3:2
AL=2AC/√3=2*8√3/√3=16

Answer 2

Answer:

16 un.

Step-by-step explanation:

In right triangle ABC:

• m∠C = 90°;
• m∠BAC = 2m∠ABC;
• BC = 24;
• AL is a bisector of angle A.

The sum of the measures of all interior angles in triangle  is always 180°, then

$m\angle ABC+m\angle BCA+m\angle BAC=180^{\circ},\\ \\m\angle ABC+90^{\circ}+2m\angle ABC=180^{\circ},\\ \\3m\angle ABC=180^{\circ}-90^{\circ}=90^{\circ},\\ \\m\angle ABC=30^{\circ},\\ \\m\angle BAC=60^{\circ}.$

In right triangle the leg that is opposite to tha angle 30° is half of the hypotenuse. This means that

$AC=\dfrac{AB}{2},\\ \\AB=2AC.$

By the Pythagorean theorem,

$AB^2=AC^2+BC^2,\\ \\(2AC)^2=24^2+AC^2,\\ \\4AC^2=576+AC^2,\\ \\3AC^2=576,\\ \\AC^2=192,\\ \\AC=8\sqrt{3}\ un.,\\ \\AB=2AC=16\sqrt{3}\ un.$

Let AL be the angle A bisector. By bisector property,

$\dfrac{AC}{AB}=\dfrac{CL}{LB},\\ \\\dfrac{8\sqrt{3}}{16\sqrt{3}}=\dfrac{CL}{24-CL},\\ \\\dfrac{1}{2}=\dfrac{CL}{24-CL},\\ \\24-CL=2CL,\\ \\3CL=24,\\ \\CL=8\ un.$

Use the Pythagorean theorem for the right triangle ACL:

$AL^2=AC^2+CL^2,\\ \\AL^2=(8\sqrt{3})^2+8^2,\\ \\AL^2=192+64=256,\\ \\AL=16\ un.$