In a triangle ABC - cos A - a -cos B - b -cos C - c If a -1-6 then the area of the triangle in square unit isA- -3 - 24B- 1 - -3C- 1 - 8D- 1 - 24

The correct option is A √3/24 Since,cosA/a = cosB/b = cosC/c ⇒cosA/k sinA=cosB/k sinB=cosC/k sinC ⇒cotA=cotB=cotC ⇒A=B=C=60^0 ⇒ΔABC is an equilateral triangle ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Basic Trigonometric Identities

In a triangle...

Question

$In a triangle ABC, \frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c} If a= \frac{1}{\sqrt{6}} then the area of the triangle (in square unit) is$

1/24

No worries! We‘ve got your back. Try BYJU‘S free classes today!

√3/24

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1/8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/√3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
√3/24

$S i n c e, \frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c} \Rightarrow \frac{c o s A}{k s i n A} = \frac{c o s B}{k s i n B} = \frac{c o s C}{k s i n C} \Rightarrow c o t A = c o t B = c o t C \Rightarrow A = B = C = 60^{0} \Rightarrow Δ ABC is an equilateral triangle ∴ Δ = \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times \frac{1}{6} = \frac{\sqrt{3}}{24} s q . u n i t$

Suggest Corrections

Similar questions

Q. In a triangle ABC, if

\frac{c o s A}{a} = \frac{cos B}{b} = \frac{cos C}{c}

and

a = \frac{1}{\sqrt{6}}

then the area of the triangle in square units is equal to?

In a triangle ABC, if $\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}$ and a = 2, then its area is

Q. If ABC is a triangle, then the vectors

(- 1, cos C, cos B)

(cos C, - 1, cos A)

and

(cos B, cos C, - 1)

are

Q. In any triangle ABC,

\frac{1 - cos A + cos B + cos C}{1 - cos C + cos A + cos B} = \frac{tan (A / 2)}{tan (C / 2)}

Q. If A, B, C be the angles of a triangle, then prove that

∣ ∣ ∣ ∣ \begin{matrix} - 1 & c o s C & c o s B c o s C & - 1 & c o s A c o s B & c o s A & - 1 \end{matrix} ∣ ∣ ∣ ∣

= 0

Join BYJU'S Learning Program

In a triangle ABC,cosAa=cosBb=cosCc If a=1√6 then the area of the triangle (in square unit) is

The correct option is B √3/24 Since,cosAa=cosBb=cosCc⇒cosAksinA=cosBksinB=cosCksinC⇒cotA=cotB=cotC⇒A=B=C=600⇒ΔABC is an equilateral triangle∴Δ=√34a2=√34×16=√324sq.unit

$In a triangle ABC, \frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c} If a= \frac{1}{\sqrt{6}} then the area of the triangle (in square unit) is$