If (a+b) cos C + (b+c) cos A + (c+a) cos B = | vedclass

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

Question

(B) Given the equation: $(a+b) \cos C + (b+c) \cos A + (c+a) \cos B = 72$ Expanding the terms: $a \cos C + b \cos C + b \cos A + c \cos A + c \cos B +

Vedclass · Accepted Answer

$216 	ext{ sq. units}$

Vedclass · Answer

(B) Given the equation: $(a+b) \cos C + (b+c) \cos A + (c+a) \cos B = 72$ Expanding the terms: $a \cos C + b \cos C + b \cos A + c \cos A + c \cos B + a \cos B = 72$ Rearranging the terms: $(a \cos C + c \cos A) + (b \cos A + a \cos B) + (b \cos C + c \cos B) = 72$ Using the projection formula: $b = c \cos A + a \cos C$,$c = a \cos B + b \cos A$,and $a = b \cos C + c \cos B$ Substituting these into the equation: $b + c + a = 72$ Given $a = 18$ and $b = 24$: $18 + 24 + c = 72$ $\Rightarrow 42 + c = 72$ $\Rightarrow c = 30$ The semi-perimeter $s = \frac{a+b+c}{2} = \frac{72}{2} = 36$ Using Heron's formula: $	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ $	ext{Area} = \sqrt{36(36-18)(36-24)(36-30)} = \sqrt{36 	imes 18 	imes 12 	imes 6}$ $	ext{Area} = \sqrt{36 	imes 1296} = 6 	imes 36 = 216 	ext{ sq. units}$

If $(a+b) \cos C + (b+c) \cos A + (c+a) \cos B = 72$ and if $a = 18, b = 24$,then the area of the triangle $ABC$ is

Similar Questions

In $\triangle ABC$,with usual notations,if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$,then $m \angle C$ is equal to

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

Which of the following sets of data does not uniquely determine an acute-angled $\Delta ABC$ ($R$ = circum-radius)?

If in a $\triangle ABC$, $r_1 < r_2 < r_3$, then:

If the angles of a triangle are in the ratio of $2 : 3 : 7$,then the sides are in the ratio of

Vedclass Test Series

Exam Paper Generator

Online Exam Module