If a = 2, b = 3, c = 5 in Delta ABC,then C = | vedclass

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

Question

(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the given values: $\cos C = \frac{2^2 + 3^2 - 5^2}{2(2)(3)}$.$\cos C

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the given values: $\cos C = \frac{2^2 + 3^2 - 5^2}{2(2)(3)}$.$\cos C = \frac{4 + 9 - 25}{12} = \frac{13 - 25}{12} = \frac{-12}{12} = -1$.Since $\cos C = -1$,we have $C = 180^{\circ}$ or $\pi$ radians.However,the sum of two sides of a triangle must be greater than the third side $(a + b > c)$.Here,$2 + 3 = 5$,which means $a + b = c$.This implies that the points $A, B, C$ are collinear and do not form a triangle.Therefore,the correct option is $D$.

If $a = 2, b = 3, c = 5$ in $\Delta ABC$,then $C = $

Similar Questions

In a triangle $ABC$,if $B = 3C$,then the values of $\sqrt{\frac{b + c}{4c}}$ and $\frac{b - c}{2c}$ are

In any triangle $ABC$,$a \cot A + b \cot B + c \cot C = $

In a triangle $ABC$,$a = 5, b = 7$ and $\sin A = \frac{3}{4}$,how many such triangles are possible?

In $\triangle ABC$,if $\angle A = 90^{\circ}$,then $2(r+R) = $

With the usual notation,in $\Delta ABC$,if $\angle A + \angle B = 120^{\circ}$ and $a : b = (\sqrt{3} + 1) : (\sqrt{3} - 1)$,then the ratio $\angle A : \angle B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module