If in a - ABC - cos A - a -cos B - b -cos C - C- then what can be said about the triangle-A- Isoceles triangleB- Right angled triangleC- Equilateral triangleD- Acute triangle

The correct option is C Equilateral triangleSince rule says that = a/sinA = b/sinB = c/sinC = k Therefore, a=k(sinA), b=k(sinB) and c=k(sinC) We can write cos ...

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Quantitative Aptitude

Triangles

If in a Δ ABC...

Question

If in a $Δ$ ABC, $\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}$ , then what can be said about the triangle?

Isoceles triangle

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Right angled triangle

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Equilateral triangle

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Acute triangle

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Solution

The correct option is C
Equilateral triangle

Since rule says that = $\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = k$
Therefore, a=k(sinA), b=k(sinB) and c=k(sinC)
We can write $\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}$ as
$\frac{c o s A}{k s i n A} = \frac{c o s B}{k s i n B} = \frac{c o s C}{k s i n C}$ = cot A=cot B=cot C
$\Rightarrow$ A=B=C(Equilateral Triangle).

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If in a ΔABC, cosAa=cosBb=cosCc, then what can be said about the triangle?

The correct option is C Equilateral triangleSince rule says that = asinA=bsinB=csinC=k Therefore, a=k(sinA), b=k(sinB) and c=k(sinC) We can write cosAa=cosBb=cosCc as cosAksinA=cosBksinB=cosCksinC = cot A=cot B=cot C ⇒ A=B=C(Equilateral Triangle).

If in a $Δ$ ABC, $\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}$ , then what can be said about the triangle?