In a triangle ABC, the perpendicular AD from point A, on the side BC meets the side BC at point D. If BD = 8 cm, DC = 2 cm and AD = 4 cm. Then, angle BAC is ________

$60^{\circ}$

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$90^{\circ}$

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$30^{\circ}$

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$45^{\circ}$

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Solution

The correct option is B
$90^{\circ}$

In right angled triangle ACD,

$A C^{2} = C D^{2} + A D^{2} \Rightarrow A C^{2} = 2^{2} + 4^{2} = 20$

In right angled triangle ABD,

$A B^{2} = A D^{2} + D B^{2} \Rightarrow A B^{2} = 4^{2} + 8^{2} = 80 c m$

$N o w B C = B D + D C = 8 + 2 = 10 c m We can observe that B C^{2} = 10^{2} = 100 and A B^{2} + A C^{2} = 80 + 20 = 100 Hence B C^{2} = A C^{2} + A B^{2} So from converse of pythagoras theorem we have ∠ B A C = 90^{\circ}$

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In a triangle ABC, the perpendicular AD from point A, on the side BC meets the side BC at point D. If BD = 8 cm, DC = 2 cm and AD = 4 cm. Then, angle BAC is ________

The correct option is B 90∘ In right angled triangle ACD, AC2=CD2+AD2⇒AC2=22+42=20 In right angled triangle ABD, AB2=AD2+DB2⇒ AB2=42+82=80 cm Now BC=BD+DC=8+2=10 cmWe can observe that BC2=102=100and AB2+AC2=80+20=100Hence BC2=AC2+AB2So from converse of pythagoras theorem we have∠BAC=90∘