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JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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In a triangle ABC,cos⁡Aa=cos⁡Bb=cos⁡CcABC,\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c}. if a=16a = \frac{1}{{\sqrt 6 }}, then the area of the triangle (in square unit) is
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Correct option: B
3/24\sqrt 3 /24
Option B (3/24\sqrt 3 /24) is correct because if cos⁡Aa=cos⁡Bb=cos⁡Cc\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c}, then the triangle is equilateral, and the area can be calculated using the side length a=16a = \frac{1}{{\sqrt 6 }}.

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If p1,p2,p3{p_1},{p_2},{p_3} are altitudes of a Δ  ABC\Delta \;ABC drawn from the vertices A,B,CA,B,C and Δ\Delta the area of the triangle, then p1−2+p2−2+p3−2p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2} is equal to If the sides of the triangles are 5k,  6k,  5k5k,\;6k,\;5k and radius of incircle is 6, the value of kk is equal to In a Δ ABC\Delta \,ABC, if a=(b−c)sec⁡θa = \left( {b - c} \right)\sec \theta, then 2 bcb−c  sin⁡  A2=\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} = In Δ  ABC,b+ca\Delta \;ABC,\frac{{b + c}}{a} is equal to If Δ=a2−(b−c)2\Delta = {a^2} - {\left( {b - c} \right)^2}, where Δ\Delta is the area of Δ  ABC\Delta \;ABC, then tan⁡A\tan A is equal to If AA is the area and 2s2 s is the sum of the sides of a triangle, then
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