*In the given figure, find the value of ∠BAE and ∠BAC*
1️⃣ ∠BAE = 132° and ∠BAC = 48°
2️⃣ ∠BAE = 101° and ∠BAC = 79°
3️⃣ ∠BAE = 127° and ∠BAC = 53°
4️⃣ ∠BAE = 127° and ∠BAC = 48°
Answers
Answer:
we know that ,
/_BCA + /_DCA = 180° ( LINEAR PAIR )
/_BCA + 132° = 180°
/_BCA = 180° - 132°
/_BCA = 48°
AND WE ALSO KNOW THAT
FROM ANGLE SUM PROPERTY OF A TRIANGLE
/_ A + /_ B + /_ C = 180°
/_ A + 53° + 48° = 180°
/_ A + 101° = 180°
/_ A = 180° - 101°
/_ A = 79°
SO
/_ BAC = 79°
AND WE KNOW THAT
/_ BAC + /_ BAE = 180° ( LINEAR PAIR )
79° + /_ BAE = 180°
/_ BAE = 180° - 79°
/_ BAE = 101°
FINAL ANSWER IS
/_BAE = 101° AND
/_ BAC 79°
option "2" is the correct answer
Answer:
Option (2) is the correct answer.
Step-by-step explanation:
Here,
< ACB+<ACD= 180°
=> <ACB+132°=180°
=> <ACB=180°-132°
=> <ACB=48°
Again,
<ABC + <ACB + <BAC = 180°
=>53° + 48° + <BAC= 180°
=>101°+ <BAC= 180°
=><BAC= 180° - 101°
=> <BAC= 79°
Again,
<BAE + <BAC = 180°
=> <BAE + 79° = 180°
=> <BAE = 180° - 79°
=> <BAE = 101°
Therefore,
(2) <BAE = 101° and <BAC= 79°