In a triangle ABC,the mid-point of BC is D. I | vedclass

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

Question

(C) In $	riangle ACD$,since $AD \perp AC$,$\angle DAC = 90^\circ$. Thus,$\cos C = \frac{AC}{CD} = \frac{b}{a/2} = \frac{2b}{a}$.In $	riangle ABC$,by

Vedclass · Accepted Answer

$\frac{2(c^2-a^2)}{3ac}$

Vedclass · Answer

(C) In $	riangle ACD$,since $AD \perp AC$,$\angle DAC = 90^\circ$. Thus,$\cos C = \frac{AC}{CD} = \frac{b}{a/2} = \frac{2b}{a}$.In $	riangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Equating the two expressions for $\cos C$:$\frac{2b}{a} = \frac{a^2+b^2-c^2}{2ab}$$4b^2 = a^2+b^2-c^2$$3b^2 = a^2-c^2 \implies b^2 = \frac{a^2-c^2}{3}$.Now,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.$\cos A \cos C = \left(\frac{b^2+c^2-a^2}{2bc}ight) \left(\frac{2b}{a}ight) = \frac{b^2+c^2-a^2}{ac}$.Substituting $b^2 = \frac{a^2-c^2}{3}$:$\cos A \cos C = \frac{\frac{a^2-c^2}{3} + c^2 - a^2}{ac} = \frac{a^2-c^2+3c^2-3a^2}{3ac} = \frac{2c^2-2a^2}{3ac} = \frac{2(c^2-a^2)}{3ac}$.

In a $\triangle ABC$,the mid-point of $BC$ is $D$. If $AD$ is perpendicular to $AC$,then $\cos A \cos C=$

Similar Questions

If in a triangle $ABC$, $\sin A, \sin B, \sin C$ are in $A.P.$, then

With usual notations,in a triangle $ABC$,if $\theta$ is any real number,then $a \cos (B-\theta) + b \cos (A+\theta)$ is equal to:

The ratios of the sides in a triangle are $5:12:13$ and its area is $270 \text{ cm}^2$. The sides of the triangle in $\text{cm}$ are

The area of an isosceles triangle is $9 \, cm^2$. If the equal sides are $6 \, cm$ in length,the angle between them is .... $^\circ$.

In a triangle $ABC$,with usual notation,$\angle B = \pi/3$ and $\angle C = \pi/4$. If $D$ divides $BC$ internally in the ratio $1:3$,find the value of $\frac{\sin \angle BAD}{\sin \angle CAD}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module