By Similar Triangles,
Angle ABC = Angle BXC
Angle ACB = Angle XCB
Angle BAC = Angle XBC
XC=1/4*AC
Solve for AC in AC/6=6/XC =
Hope that helps.
In straight angle AXC,
angle BXC = 180 -70 = 110 degrees
In triangle BXC,
Let angle XBC = theta
(theta +angle C) = 180 -110 = 70 deg
In big triangle ABC
(angle A +angle C) = 180 -110 = 70 deg also
So,
theta +angle C = angle A +angle C
Hence,
angle A = theta
---------------
If AX /XC = 3,
then side AC is divided into 4 equal parts.
Let w = length of AC,
So, XC = w/4
In triangle BXC, by Law of Sines,
BC /sin(angle BXC) = XC /sin(angle XBC)
6 /sin(110deg) = (w/4) /sin(theta)
Cross multiply,
6sin(theta) = (w/4)sin(110deg)
sin(theta) = w*sin(110deg) /24 ------**
In triangle ABC, by Law of Sines,
AC /sin(angle ABC) = BC /sin(angle A)
w /sin(110deg) = 6 /sin(theta)
Cross multiply,
w*sin(theta) = 6sin(110deg)
sin(theta) = 6sin(110deg) /w -----**
sin(theta) = sin(theta)
w*sin(110deg) /24 = 6sin(110deg) /w
w/24 = 6/w
w^2 = 6*24
w = sqrt(144) = 12
Therefore, AC = 12 cm ----------answer.
Hello, googa!
A slight variation of ticbol's solution . . .
In is a point on such that
and cm.
Find the length ofCode:B o * ** * * * * * β* 6 * * * * * * * * * * β 70° * α * o * * * * * * * o * o A 3a X a C
We have: .
Let:
Consider and
. . They share
Since , then
. . They have equal angles: .
Hence, their third angles are equal: .
In .[1]
In .[2]
Equate [1] and [2]: .
Therefore: .