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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $........^{\circ}$

Similar Questions

In a triangle $ABC$, if $a=3, b=4$ and $\sin A=\frac{3}{4}$, then $\angle CBA = (\text{in } ^{\circ})?$

In a triangle $ABC$ with usual notations,if $a:b:c = 7:8:9$,then $\cos A : \cos B : \cos C =$

In a $\Delta ABC$,if $A = 30^o$,$b = 2$,and $c = \sqrt{3} + 1$,then $\frac{C - B}{2} = $ ....$^o$

In a triangle $ABC$,with usual notations,if $a=4, b=8, \angle C=60^{\circ}$,then the value of $\angle B$ and the ratio $\cos A : \cos C$ respectively are,

In $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module