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ΔABC sama sisi = 6
buat titik D pada AB, D = titik tengah AB
AD= DB = 1/2 (6) = 3
CD tegak lurus AB, maka CD = √(CB²- DB²)=√(6²- 3²)= √27= 3√3
P titk Berat ABC , CP = 2/3 CD = 2/3.3√3 = 2√3
TP tegak lurus CD
TP² = TC² - CD²
TP = √(6√3)²- (2√3)² = √(108 - 12)= √96 = 4√6
sudut (TC, ABC) = α
Tan α = TP/CP = 4√6 / (2√3) = (2√6)/(√3) = 2√2