If Delta = a^2 - (b - c)^2 is the area of the | vedclass

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

Question

(B) Given $\Delta = a^2 - (b - c)^2$. Using the identity $x^2 - y^2 = (x - y)(x + y)$,we have $\Delta = (a - (b - c))(a + (b - c)) = (a - b + c)(a + b

Vedclass · Accepted Answer

$\frac{8}{15}$

Vedclass · Answer

(B) Given $\Delta = a^2 - (b - c)^2$. Using the identity $x^2 - y^2 = (x - y)(x + y)$,we have $\Delta = (a - (b - c))(a + (b - c)) = (a - b + c)(a + b - c)$. Since $2s = a + b + c$,we have $a + b - c = 2s - 2c$ and $a - b + c = 2s - 2b$. Thus,$\Delta = (2s - 2b)(2s - 2c) = 4(s - b)(s - c)$. We know that $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$. Equating the two expressions: $\sqrt{s(s - a)(s - b)(s - c)} = 4(s - b)(s - c)$. Dividing both sides by $\sqrt{(s - b)(s - c)}$,we get $\sqrt{s(s - a)} = 4\sqrt{(s - b)(s - c)}$. Therefore,$\sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{1}{4}$. Since $	an \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$,we have $	an \frac{A}{2} = \frac{1}{4}$. Using the formula $	an A = \frac{2 	an \frac{A}{2}}{1 - 	an^2 \frac{A}{2}}$,we get $	an A = \frac{2(1/4)}{1 - (1/4)^2} = \frac{1/2}{1 - 1/16} = \frac{1/2}{15/16} = \frac{8}{15}$.

If $\Delta = a^2 - (b - c)^2$ is the area of the $\triangle ABC$,then $\tan A$ is equal to

Similar Questions

Acidity of diprotic acids in aqueous solutions increases in the order

An aircraft executes a horizontal loop with a speed of $150 \, m/s$ with its wings banked at an angle of $12^\circ$. The radius of the loop is .......... $km$. $(g = 10 \, m/s^2)$

Which of the following is an emulsifier?

The binding energy of deuteron is $2.2\,MeV$ and that of $_2^4He$ is $28\,MeV$. If two deuterons are fused to form one $_2^4He$ nucleus, then the energy released is ......... $MeV$.

At a certain temperature,the equilibrium constant $K_c$ is $0.25$ for the reaction:
$A_{2(g)} + B_{2(g)} \rightleftharpoons C_{2(g)} + D_{2(g)}$
If we take $1 \ mol$ of each of the four gases in a $10 \ L$ container,what would be the equilibrium concentration of $A_{2(g)}$ (in $M$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module