Question

1. right δabc

bd bisects ∠abc
m∠a = 36°
find m∠bdc.

Explanation:

Step1: Find ∠ABC in right ΔABC

In right triangle \( \triangle ABC \), \( \angle B = 90^\circ \) (right angle), \( \angle A = 36^\circ \). By triangle angle - sum property (\( \angle A+\angle B+\angle C = 180^\circ \)), we can find \( \angle C \), but first we know \( \angle ABC=90^\circ \) (given it's a right triangle at B).

Step2: Find ∠ABD and ∠DBC

Since \( BD \) bisects \( \angle ABC \), and \( \angle ABC = 90^\circ \), then \( \angle ABD=\angle DBC=\frac{1}{2}\angle ABC \). So \( \angle DBC=\frac{90^\circ}{2} = 45^\circ \).

Step3: Find ∠C in ΔABC

Using the triangle angle - sum formula \( \angle A+\angle B+\angle C=180^\circ \), substitute \( \angle A = 36^\circ \), \( \angle B = 90^\circ \). Then \( 36^\circ+90^\circ+\angle C = 180^\circ \), so \( \angle C=180^\circ-(36^\circ + 90^\circ)=54^\circ \).

Step4: Find ∠BDC in ΔBDC

In \( \triangle BDC \), using the triangle angle - sum property (\( \angle DBC+\angle C+\angle BDC = 180^\circ \)). We know \( \angle DBC = 45^\circ \), \( \angle C = 54^\circ \). So \( \angle BDC=180^\circ-(45^\circ + 54^\circ)=180^\circ - 99^\circ = 81^\circ \).

Answer:

\( 81^\circ \)