Question

IF ONE OF ANGLE OF TRIANGLE IS 100 DEGREE FIND THE ANGLE BETWEEN THE BISECTOR OF OTHER TWO ANGLE?

Answer 1

Consider a △ABC,such that ∠BAC=100

∘

and bisectors of ∠B and ∠C meet at O.

To find: ∠BOC

Now, in △ABC,

∠BAC+∠ABC+∠ACB=180

100+∠ABC+∠ACB=180 (Angle sum property)

∠ABC+∠ACB=80

1/2 (∠ABC+∠ACB)=40

∠OBC+∠OCB=40 (OB and OC bisect ∠ABC and ∠ACB)

Now, in △OBC,

∠OBC+∠OCB+∠BOC=180

40+∠BOC=180

∠BOC=140

∘

.

Answer 2

Given:

In a triangle one angle is $\[100^\circ \]$

Solution:

Know that, The angle between the bisector of the other two angle$\[ = 90^\circ + \frac{{\angle A}}{2}\]$

Assume that in triangle ABC, the $\[\angle A = 100^\circ \]$,

Therefore, the angle between the bisector of the other two angle is,

$\[ = 90^\circ + \frac{{\angle A}}{2}\]$

$\[ = 90^\circ + \frac{{100^\circ }}{2}\]$

$\[\begin{array}{l} = 90^\circ + 50^\circ \\ = 140^\circ \end{array}\]$

Hence, the angle between the bisector of the other two angle is $\[140^\circ \]$.