If the sides of a triangle are 3, 5, 7,then | vedclass

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

Question

(B) Let the sides be $a = 3$,$b = 5$,and $c = 7$.To determine the nature of the triangle,we calculate the cosine of the largest angle,which is opposit

Vedclass · Accepted Answer

One angle is obtuse

Vedclass · Answer

(B) Let the sides be $a = 3$,$b = 5$,and $c = 7$.To determine the nature of the triangle,we calculate the cosine of the largest angle,which is opposite the longest side $(c = 7)$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2 	imes 3 	imes 5} = \frac{9 + 25 - 49}{30} = \frac{34 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$.Since $\cos C = -\frac{1}{2}$,the angle $C = 120^{\circ}$.Because $C > 90^{\circ}$,the triangle has one obtuse angle.

If the sides of a triangle are $3, 5, 7$,then

Similar Questions

In a $\triangle ABC$,if $a=6$,$b=5$,and $c=4$,then find the value of $\cos 2A$.

If ${p_1}, {p_2}, {p_3}$ are altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively and $\Delta$ is the area of the triangle,then ${p_1}^{-2} + {p_2}^{-2} + {p_3}^{-2}$ is equal to

In $\triangle ABC$,if $r_1=8, r_2=12$ and $r_3=24$,then the ordered triple $(a, b, c) =$

In any $\triangle ABC$,the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

In a $\Delta ABC$,$\angle A = 30^\circ$ and $a = 8 \text{ cm}$,then the distance of the orthocentre from vertex $A$ is equal to (with usual notations).

Vedclass Test Series

Exam Paper Generator

Online Exam Module