In a triangle ABC,the altitudes from B and C | vedclass

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(B) Let $BE$ and $CF$ be the altitudes from $B$ and $C$ to the sides $AC$ and $AB$ respectively.Given that $BE \geq AC$ and $CF \geq AB$.In $	riangle

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let $BE$ and $CF$ be the altitudes from $B$ and $C$ to the sides $AC$ and $AB$ respectively.Given that $BE \geq AC$ and $CF \geq AB$.In $	riangle ABE$,$\sin A = \frac{BE}{AB} \implies BE = AB \sin A$.Since $BE \geq AC$,we have $AB \sin A \geq AC \dots (i)$.In $	riangle ACF$,$\sin A = \frac{CF}{AC} \implies CF = AC \sin A$.Since $CF \geq AB$,we have $AC \sin A \geq AB \dots (ii)$.Multiplying $(i)$ and $(ii)$,we get $(AB \cdot AC) \sin^2 A \geq (AB \cdot AC) \implies \sin^2 A \geq 1$.Since $\sin A \leq 1$,this implies $\sin^2 A = 1$,so $\sin A = 1$,which means $A = 90^{\circ}$.Substituting $A = 90^{\circ}$ into $(i)$ and $(ii)$,we get $AB \geq AC$ and $AC \geq AB$,so $AB = AC$.Thus,the angles of the triangle are $90^{\circ}, 45^{\circ}, 45^{\circ}$.

In a $\triangle ABC$,the altitudes from $B$ and $C$ to the opposite sides are not shorter than their respective opposite sides. Then,one of the angles of $\triangle ABC$ is $........^{\circ}$

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