Dear student, Given,∠BAC=45o ,∠ACD=105o and ∠ABC=xNow,∠ACD+∠ACB=180o (∵BCD is a straight line)⇒∠ACB=180o-∠ACD⇒∠ACB=180o-105o⇒∠ACB=75oNow,since ABC is a triangle,Sum of internal angles =180o⇒∠ACB+∠ABC+∠BAC=180o⇒75o+x+45o=180o⇒x+120o=180o⇒x=180o-120o⇒x=60o Hope this information will clear your doubts about topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

Find the x

Dear student,

G i v e n, ∠ B A C = 45^{o}, ∠ A C D = 105^{o} a n d ∠ A B C = x N o w, ∠ A C D + ∠ A C B = 180^{o} (∵ B C D i s a s t r a i g h t l i n e) \Rightarrow ∠ A C B = 180^{o} - ∠ A C D \Rightarrow ∠ A C B = 180^{o} - 105^{o} \Rightarrow ∠ A C B = 75^{o} N o w, \sin c e A B C i s a t r i a n g l e, S u m o f i n t e r n a l a n g l e s = 180^{o} \Rightarrow ∠ A C B + ∠ A B C + ∠ B A C = 180^{o} \Rightarrow 75^{o} + x + 45^{o} = 180^{o} \Rightarrow x + 120^{o} = 180^{o} \Rightarrow x = 180^{o} - 120^{o} \Rightarrow x = 60^{o}

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

-2

View Full Answer