Advertisements
Advertisements
Question
In ΔABC, `cos"A"/"a" = cos"B"/"b" cos"C"/"c"`. If a = `1/sqrt(6)`, then the area of the triangle is ______.
Options
`1/8` sq. units
`1/(8sqrt(3)` sq. units
`1/24` sq. units
`1/(24sqrt(3)` sq. units
MCQ
Fill in the Blanks
Solution
In ΔABC, `cos"A"/"a" = cos"B"/"b" cos"C"/"c"`. If a = `1/sqrt(6)`, then the area of the triangle is `1/(8sqrt(3)` sq. units.
Explanation:
Given, `cos"A"/"a" = cos"B"/"b" cos"C"/"c"`
∴ `cos"A"/("k" sin "A") = cos"B"/("k" sin "B")= cos"C"/("k" sin "C")` .......[By sine rule]
⇒ cot A = cot B = cot C
∴ ΔABC is equilateral.
∴ Area of the equilateral triangle = `sqrt(3)/4` (side)2
= `sqrt(3)/4 (1/sqrt(6))^2`
= `sqrt(3)/24`
= `1/(8sqrt(3))` sq. unis
shaalaa.com
Is there an error in this question or solution?
