Question

A region is a shape of a triangle ABC with two sides a=75 m. b = 130 m, and C=118°
Find the approximate area of the triangular region.​

Answer 1

Answer:

you may use this formula,√s(s-a)(s-b)(s-c)

wher a,b,c are sides and s means

(a+b+c)/2

Answer 2

The approximate area of the triangular region=$4304m^2$

Step-by-step explanation:

a=75 m,b=130 m

$C=118^{\circ}$

Cosine law:$c^2=a^2+b^2-2abcosC$

By using the cosine law

$c=\sqrt{(75)^2+(130)^2-2(75)(130)cos(118)}$

$c=\sqrt{(75)^2+(130)^2+2(75)(130)(0.469)}=178 m$

$Heron's formula:\sqrt{s(s-a)(s-b)(s-c)}$

Where s= Half perimeter

Half perimeter of triangular shape ABC=$\frac{a+b+c}{2}=\frac{75+130+178}{2}=191.5 m$

By using Herone's formula

Area of triangular shape ABC=$\sqrt{191.5(191.5-75)(191.5-130)(191.5-178)}$

Area of triangular shape ABC=$4303.8m^2\approx 4304m^2$

Hence, the approximate area of the triangular region=$4304m^2$

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