In triangle ABC,if a=4, b=3, c=2,then 2(a-b c | vedclass

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

Question

(D) Given $a=4, b=3, c=2$ in $	riangle ABC$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{16+9-4}{2 	imes 4 	imes 3} = \frac

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given $a=4, b=3, c=2$ in $	riangle ABC$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{16+9-4}{2 	imes 4 	imes 3} = \frac{21}{24} = \frac{7}{8}$. $\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{16+4-9}{2 	imes 4 	imes 2} = \frac{11}{16} \implies \sec B = \frac{16}{11}$. Now,substitute these values into the expression: $2(a-b \cos C)(a-c \sec B) = 2(4 - 3 	imes \frac{7}{8})(4 - 2 	imes \frac{16}{11})$ $= 2(4 - \frac{21}{8})(4 - \frac{32}{11})$ $= 2(\frac{32-21}{8})(\frac{44-32}{11})$ $= 2(\frac{11}{8})(\frac{12}{11}) = 2 	imes \frac{12}{8} = 2 	imes \frac{3}{2} = 3$.

In triangle $ABC$,if $a=4, b=3, c=2$,then $2(a-b \cos C)(a-c \sec B) = $

Similar Questions

In triangle $ABC$,if $A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$,then $c =$

If in a triangle $ABC$,$\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6}$,then the value of $\cos A + \cos B + \cos C$ is equal to

In a $\triangle ABC$,if $(a+b+c)(b+c-a) = \lambda bc$,then which of the following is true?

In $\Delta ABC,$ if $a = 16, b = 24$ and $c = 20,$ then $\cos \frac{B}{2} = $

If in $\triangle ABC$,with usual notations,$a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module