Question

An Isosceles triangle ABC has its vertices on a if /AB/=13cm,/BC/=13cm and /AC/= 10cm, calculate

Answer 1

Explanation:

you didn't provide enough information for the question but I'll find the area, perimeter,height, and all the angles in it for you,

Area=
`\sqrt{(s(s-a)(s-b)(s-c))`

where s= a+b+c/2

36/2

=18cm=S

Area=
`\sqrt{(18(18-13)(18-13)(18-10))`

`\sqrt{3600`

=60cm^2=Area

perimeter=10+13+13=36

angles=
`cos(C) = (a^2 + b^2 - c^2) / 2ab`

=cos(C)=(13^2+13^2-10^2)/2(13)(13)

cos (C)=0.704142011

(C)=
`cos^-1`(0.704142011)

(C)=45.24 degrees

(A)=

Sin(A)/13=Sin(45.24)/10

(A)=
`sin^-1`(0.34172167)

(A)=19.98 degrees

(B)=19.98+45.24

(B)=180-65.22

(B)=114.78 degrees

the height can be found by dividing the base (10) into two parts

5 and 5 and draw a line from AC to B (perpendicular)

13^2-5^2=144

c=12cm=height