Question

Geometry help please?
In an isosceles triangle ABC, AB = AC

A = 20*

Point P is taken on side AC such that AP = BC.
Find PBC.

Answer 1

$$BC=2AC\sin(10^\circ)$$
$$PC=AC-2AC\sin(10^\circ)=AC[1-2\sin(10^\circ)]$$
from sine rule to PCB triangle,
$$\frac{PC}{\sin \alpha}=\frac{BC}{\sin[180-\alpha-80]}$$

now solve for alpha

Answer 2

u know it then plz tell it

Answer 3

why so much suspense

Answer 4

in triangle abc
ab=ac
ap=bc
<a=20
to find <pbc

Answer 5

after this what should i do

Answer 6

and i didnt understan what badhi told

Answer 7

explain plz

Answer 8

hello

Answer 9

oh

Answer 10

so u have any solution

Answer 11

no i dont have the answer

Answer 12

i never said that

Answer 13

oh then pls tell

Answer 14

i forgot to put the q mark

Answer 15

have any idea?

Answer 16

do we have to use trigo here

Answer 17

rotation

Answer 18

any other method that we can use here

Answer 19

ok

Answer 20

mertsj

Answer 21

have any ideas

Answer 22

Do you want to talk or do you want to work on this problem?

Answer 23

no uh actually i have to go now
so we will meet tomorrow same time sama place

Answer 24

please tell me the sol after wards

Answer 25

Bye bye

Answer 26

will meet tomorrow?

Answer 27

We'll see. Maybe later today. I'll post a solution if I find one.

Answer 28

ok

Answer 29

bye

Answer 30

still not got

Answer 31

is that the angle of A you have given or a length??

Answer 32

angle

Answer 33

HOW CAN ANGLE B AND ANGLE C BE 170

Answer 34

cosine rule
\[A ^{2}=B ^{2}+C ^{2}-2BCcosA ^{*}\]

Answer 35

BUT ANGLE B AND ANGLE C ARE NOT 170

Answer 36

AS ANGLE IN A TRIANGLE IS 180

Answer 37

i am sorry it is rather