Question

[Chi Chi]

Cho tam giác ABC có góc A = 64 độ. Đường cao AH; góc BAH = 2 x góc CAH. Tính góc B; góc C

Answer 1

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Ta có : BAH = 2CAH (gt)

Mà BAH + CAH = BAC = 64° 

=> CAH + 2CAH= 64° 

=> 3CAH = 64° 

=> CAH = \(\frac{64}{3}\)

=> BAH = 64 - \(\frac{64}{3}\)=\(\frac{128}{3}\)

=> C = 180 - 90 - \(\frac{64}{3}\)=\(\frac{206}{3}\)

=> B = 180 - 90 - \(\frac{128}{3}\)= \(\frac{142}{3}\)