In a - ABC- if a -2- b -3 and c -5- show that its area is 1-2-6 sq-units-

The area of a triangle ABC is given by Δ=1/2ab sin C cos C=a^2+b^2 c^2/2abc =2+3 5/2√(6)=0 sin C=√(1 cos^2 C=1) Therefore, Δ=1/2ab sin C =1/2 √(6) ...

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Standard XII

Mathematics

Cosine Rule

In a Δ ABC, i...

Question

$I n a Δ A B C, i f a = \sqrt{2}, b = \sqrt{3} a n d c = \sqrt{5}, show that its area is \frac{1}{2} \sqrt{6} s q . u n i t s .$

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Solution

$The area of a triangle ABC is given by Δ = \frac{1}{2} a b s i n C c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b c} = \frac{2 + 3 - 5}{2 \sqrt{6}} = 0 s i n C = \sqrt{1 - c o s^{2} C = 1} Therefore, Δ = \frac{1}{2} a b s i n C = \frac{1}{2} \sqrt{6}$

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In a ΔABC, if a=√2, b=√3 and c=√5, show that its area is 12√6 sq.units.

The area of a triangle ABC is given byΔ=12ab sin Ccos C=a2+b2−c22abc=2+3−52√6=0sin C=√1−cos2 C=1Therefore,Δ=12ab sin C=12 √6

$I n a Δ A B C, i f a = \sqrt{2}, b = \sqrt{3} a n d c = \sqrt{5}, show that its area is \frac{1}{2} \sqrt{6} s q . u n i t s .$

$The area of a triangle ABC is given by Δ = \frac{1}{2} a b s i n C c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b c} = \frac{2 + 3 - 5}{2 \sqrt{6}} = 0 s i n C = \sqrt{1 - c o s^{2} C = 1} Therefore, Δ = \frac{1}{2} a b s i n C = \frac{1}{2} \sqrt{6}$