Question

HELP ME EXTRA POINTS

Answer 1

Given: CAB=CBA; AC=BC; H=90; HCA=80
To find: HAB=?

In Triangle HAC,
AHC=90 (given)
HCA=80 (given)
HCA+AHC+HAC=180 (ASP of a Triangle)
= 80+90+HAC=180
AHC=10

Now,
HCA+ACB=180 (Linear Pair)
= 80+ACB=180
ACB=100

In Triangle ACB,
CAB=CBA (given)
So let them both be x
= x+x+ACB=180 (ASP of triangles)
= 2x=180-100
= x=80/2
x= 40
Therefore, CAB=CBA=40

HAB=AHC+CAB
= 10+40

Thus, Angle HAB=50 degrees

Answer 2

50

Explanation:

We know that the sum of the angles of a triangle is 180

HCA + CAH + AHB = 180

80+ CAH + 90 = 180

Combine like terms

170 + HAC = 180

HAC= 180-170

HAC = 10

We also know that HCA + ACB is a straight line so it adds to 180

HCA + ACB = 180

80+ACB = 180

ACB = 180-80

ACB = 100

Since ACB is isosceles the base angles have the same measure so CAB = ABC

We know that the sum of the angles of a triangle is 180

CAB + ABC + ACB = 180

Replacing ABC with CAB

CAB + CAB + 100 = 180

2CAB + 100 =180

2 CAB = 180-100

2 CAB = 80

Divide by 2

CAB = 80/2

CAB = 40

Now find HAB

HAB = HAC + CAB

= 10 + 40

= 50