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Sign of Trigonometric Ratios in Different Quadrants

In a triangle...

Question

In a triangle ABC , BD perpendicular on AC and CE perpendicular on AB .if BD and CE intersect at o , prove that angle boc=180^°-angle a

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Q. In ∆ABC, BD is perpendicular to AC and CE is perpendicular to AB. If BD and CE intersect at O, prove that angleBOC = 180° - angle A.

In $Δ A B C, B D ⊥ A C$ and CE $⊥ A B .$ If BD and CE intersect at O , prove that $∠ B O C = 180^{\circ} - ∠ A .$

Q. In Δ ABC, BD⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° − A.

Q. In a triangle ABC, BD and CE are perpendiculars to the sides AC and AB respectively. If BD and CE intersect at I, then,

∠ B I C = 180^{\circ} - ∠ A

. State true or false

State true or false.

In a triangle ABC, BD and CE are perpendiculars to the sides AC and AB respectively. If BD and CE intersect at I, then

∠ B I C

180^{\circ}

∠ A

.

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