If H is orthocentre of triangle ABC, then AH is ? (a) b cot A (b) c cot A (c) a cot A (d) a cot B Share with your friends Share 2 Manbar Singh answered this In ∆AHB, we haveAHsin90°-A = csinA+B⇒AHcos A = csin180°-C⇒AH = c cos Asin C⇒AH = csin C ×cos A⇒AH = asin A × cos A ∵ asin A = bsin B = c sin C⇒AH = a × cos Asin A⇒AH = a cot A 13 View Full Answer
