Question

In Figure 10.24, AB = AC and ∠ACD =105°, find ∠BAC. 

 

Answer 1

Consider the given figure
We have, 

AB=AC and ∠ ACD=105 ° 

Since, 

⇒ ∠BCD=180°=straight angle  

⇒∠BCA+∠ACD=180° 

⇒∠BCA=180° -105°  ⇒ ∠BCA=75°                .......................(1)  

And also, 

ΔABC is an isosceles triangle            [∵ AB=AC] 

⇒ ∠ABC = ∠ACB    [Angles opposite to equal sides are equal ] 

From (1), we have  

∠ ACB=75⇒ ∠ ABC=∠ ACB = 75° 

And also,
Sum of interior angles of a triangle=180°  

 ⇒ ∠ ABC=∠BCA+∠CAB =180° 

 ⇒ 75° +75° +∠CAB=180°  

⇒ 150° +∠BAC=180°  ⇒∠BAC=180° -150° =30° 

∴ ∠BAC =30°