Question

Find angle ABC and angle ACB in the figure given below. ​

Answer 1

$\huge\mathfrak{\red{Answer}}$

➡ / A = 70°

➡ / B = 65°

➡ / C = 45°

Step-by-step explanation:

Given,

<A=70°

Now,

<y+135°= 180°

[sum of angles on the same line is 180°]

now,

<y= 180°-135°

<y = 45°

Also,

In /\ ABC

/ A + / B + / C = 180°

[Angle sum property]

70°+x+y =180°

=> 70° + x + 45° = 180°

=> x = 180°-115°

=> x = 65°

Answer 2

$\huge \colorbox{cyan}{given}$

==》<BAC = 70⁰

==》<ACD = 135⁰

$\huge \colorbox{red}{to \: find}$

==》<ABC = ?

==》<ACB = ?

$\huge \colorbox{green}{soluton}$

==》In triangle ABC

==》 ACD IS AN exterior angle .

$\huge \colorbox{pink}{theorem \: 2 \: says}$

==》If a side of a triangle is produced ,then the exterior angles so formed is equal to the sum of the two interior opposite angles

so ,

==》135⁰ = 70⁰ + X

==》X = 135⁰ - 70⁰

==》X = 65⁰

==》<ABC = 65⁰

$\huge \colorbox{grey}{tip}$

==》you can also do this by using linear pair of angles .

NOW,

==》<ABC + <ACB + <BAC = 180⁰

{ ANGLE SUM PROPERTY OF TRIANGLE}

==》65⁰ + <ACB + 70⁰ = 180⁰

==》 135⁰ +<ACB = 180⁰

==》 <ACB = 180⁰ - 135⁰

==》<ACB = 45⁰

$\huge \colorbox{cyan}{hence}$

⏩<ABC = 65⁰

⏩<ACB = 45⁰

⏩<BAC = 70⁰

$\huge \colorbox{purple}{note}$

==》An exterior angle of triangle is greater than either of its interior opposite angle .

==》Theorem 1 says =》The sum of interior angles of a triangle is 180⁰ .