Question

help plz 7th grade questions ​

Answer 1

Answer:

here ab||CE

and ac is transversal

bac= ace..... alternate angle

BAC= 40......GIVEN ....1

ABC=BCD...... CORRESPONDING ANGLE

ABC= 80.....GIVE ... 2

NOW IN TRAINGLE ABC

A+B+C= 180°

40+80+C=180... FROM 1AND 2

C= 180-120

C= 60

A= 40 AND B= 80 and c= 60

Step-by-step explanation:

hope it will help you... Try to follow me so that I can help you in next example

Answer 2

$\color{red}\huge{\underline{{\tt GIVEN : -}}}$

• $\tt \angle ACE=40°$
• $\tt\angle ECD=80°$
• $\tt and \: \: BA \: || \: CE$

$\color{red}\huge{\underline{{\tt SOLUTION : -}}}$

$\tt \purple{since : }$

$\tt \angle ACB ,\angle ACE \: \: and \: \: \angle ECD \: lie \: on \: \tt same \: plane.$

$\implies \tt \angle ACB + \angle ACE + \angle ECD = 180 \degree$

$\implies \tt \angle ACB + 40 \degree + 80 \degree = 180 \degree(given)$

$\implies \tt \angle ACB + 120 \degree = 180 \degree$

$\implies \tt \angle ACB = 180 \degree - 120 \degree$

$\boxed{ \orange{\therefore \tt \angle ACB = 60 \degree}}...(1)$

$\purple{ \tt since : }$

$\tt BA \: || \: CE(BA \: \: parallel \: \: to \: \: CE)$

$\tt \angle ABC=\angle ECD (corresponding \: angles)$

$\boxed{ \orange{ \tt\therefore \ \angle ABC=80 \degree}}...(2)$

$\purple{ \tt by \: \: angle \: \: sum \: \: property : }$

$\tt\angle ABC+\angle ACB+\angle BAC=180°$

$\purple{ \tt from \: \: (1) \: \: and \: (2) : }$

$\implies \tt60 \degree + 80 \degree + \angle BAC=180°$

$\implies \tt 140 \degree + \angle BAC=180°$

$\implies \tt \angle BAC=180° -140 \degree$

$\boxed{ \orange{ \tt\therefore \angle BAC=40°}}$

Therefore,

Each angle of the triangle ABC is:

• $\blue{ \tt\angle BAC=40°}$
• $\blue{ \tt\angle ABC=80°}$
• $\tt\blue{\angle ACB=60°}$

HOPE IT HELPS!!