Q:

# Given: ∆ABC, m∠C = 90° m∠BAC = 2m∠ABC BC = 24, AL −∠ bisector Find: AL Will be brainliest answer if right!!!

Accepted Solution

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