Question

BD−→− bisects ∠ABC such that m∠ABD = (4x−5)∘ and m∠DBC = (3x+2)∘.

Answer 1

Answer:

$\angle ABC=(46)^{\circ}$

Step-by-step explanation:

Given: BD bisects $\angle ABC$

$\angle ABD=(4x-5)^{\circ}\\\angle DBC=(3x+2)^{\circ}$

To find: $\angle ABC$

Solution:

A bisector of an angle divides it into angles of equal measures.

As BD bisects $\angle ABC$, $\angle ABD=\angle DBC$

Put $\angle ABD=(4x-5)^{\circ}\,,\,\angle DBC=(3x+2)^{\circ}$

$\angle ABD=\angle DBC\\(4x-5)^{\circ}=(3x+2)^{\circ}\\4x-3x=2+5\\x=7^{\circ}$

So,

$\angle ABD=(4x-5)^{\circ}=(4(7)-5)^{\circ}=(28-5)^{\circ}=(23)^{\circ}\\\angle DBC=(3x+2)^{\circ}=(3(7)+2)^{\circ}=(21+2)^{\circ}=(23)^{\circ}$

$\angle ABC=\angle ABD+\angle DBC\\=(23)^{\circ}+(23)^{\circ}\\=(46)^{\circ}$