Question

in triangle abc anglea:angleb:angle c=1:2:3 then find bc:ab:ac​

Answer 1

Answer:

1:2:√3

Step-by-step explanation:

Given--->in ΔABC

-----------

∠A:∠B:∠C = 1:2:3

To find--->

-----------

BC:AB:AC=?

Solution---->

--------------

∠A:∠B:∠C=1:2:3

let ∠A=x ,∠B=2x,∠C=3x

sum of angles of triangle are equal to 180⁰

so ∠A+∠B+∠C=180⁰

=> x +2 x +3x =180⁰

=> 6x =180⁰

180⁰

=> x=-------------

3

=> x =30⁰

so ∠A =x=30⁰

∠B=2x=2(30⁰)=60⁰

∠C=3x=3(30⁰)=90⁰

now we have sine rule

a b c

--------- = ----------=-----------

sinA sinB sinc

BC AC AB

---------= -----------=-----------

Sin30⁰ sin60⁰ sin90⁰

BC:AC:AB=sin30*:sin60*:sin90*

BC:AB:AC=sin30*:sin90*:sin60*

1 √3

BC:AB:AC= ------ : 1 : ------

2 2

multiplying ratios by 2

1 √3

BC:AB:AC=2×------: 2×1 :2×------

2 2

= 1 : 2 : √3

Hope it helps u

Thanks