In $△ A B C$ if $r_{1} = r_{2} + r_{3} + r$ . Then the triangle is

Acute angled

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

Right Angled

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

Equilateral

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

Isosceles

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

Open in App

Solution

The correct option is A Right Angled
$r_{1} = r_{2} + r_{3} + r ⟹ r_{1} - r = r_{2} + r_{3}$
$⟹ sin \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2} - sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = sin \frac{B}{2} cos \frac{A}{2} cos \frac{C}{2} - sin \frac{C}{2} cos \frac{A}{2} cos \frac{B}{2}$
$⟹ sin \frac{A}{2} (cos \frac{B}{2} cos \frac{C}{2} - sin \frac{B}{2} sin \frac{C}{2}) = cos \frac{A}{2} (sin \frac{B}{2} cos \frac{C}{2} - sin \frac{C}{2} cos \frac{B}{2})$
$⟹ sin \frac{A}{2} cos \frac{(B + C)}{2} = cos \frac{A}{2} sin \frac{(B + C)}{2}$
$⟹ tan \frac{A}{2} = tan \frac{(B + C)}{2}$

$⟹ A = B + C ⟹ 2 A = 180^{\circ} ⟹ A = 90^{\circ}$

Suggest Corrections

Similar questions

r_{1}, r_{2}, r_{3}

A B C

r

r_{1} r_{2} r_{3} - r (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})

△ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

\frac{r}{r_{1}} = \frac{r_{2}}{r_{3}},

r, r_{1}, r_{2}, r_{3}

In acute angled triangle $A B C, r + r_{1} = r_{2} + r_{3}$ and $∠ B > \frac{π}{3}$ then

A B C, \frac{r_{1} - r}{a} + \frac{r_{2} - r}{b} + \frac{r_{3} - r}{c} =

Join BYJU'S Learning Program