Question

based on the information provided below, which two statements are true?
in △dab, m∠d = 30° and m∠dab = 120°. in △abc, ab = 10 and bc = 10.
□ △abc is an obtuse triangle
□ ∠d = m∠dba
□ △dbc is a right triangle
□ db = dc
□ db = 20

Explanation:

Step1: Analyze △DAB

In △DAB, \( m\angle D = 30^\circ \), \( m\angle DAB = 120^\circ \). So \( m\angle DBA=180^\circ - 30^\circ - 120^\circ = 30^\circ \). Thus \( m\angle D = m\angle DBA \), so \( DA = BA \).

Step2: Analyze △ABC

Given \( AB = BC = 10 \), so △ABC is isosceles. \( \angle BAC = 180^\circ - 120^\circ = 60^\circ \), so △ABC is equilateral (all angles \( 60^\circ \), all sides equal), so it's not obtuse.

Step3: Analyze △DBC

Since \( DA = BA \) and \( AB = BC \), \( DA = BC \), and \( DC = DA + AC = DA + AB \) (since AC=AB in equilateral △ABC). Wait, no, let's check angles. \( \angle DAB = 120^\circ \), \( \angle BAC = 60^\circ \), so \( \angle DAC = 180^\circ \)? No, wait, D, A, C are colinear? So \( DC = DA + AC \). But from △DAB, \( \angle DBA = 30^\circ \), \( AB = BC \), \( \angle ABC = 60^\circ \), so \( \angle DBC = \angle DBA + \angle ABC = 30^\circ + 60^\circ = 90^\circ \). So △DBC is right-angled at B. Also, \( DB \): in △DAB, using sine law, \( \frac{DB}{\sin 120^\circ} = \frac{AB}{\sin 30^\circ} \), so \( DB = AB \cdot \frac{\sin 120^\circ}{\sin 30^\circ} = 10 \cdot \frac{\sqrt{3}/2}{1/2} = 10\sqrt{3} \). \( DC \): since △ABC is equilateral, \( AC = 10 \), \( DA \): in △DAB, \( DA = AB = 10 \) (since \( \angle D = \angle DBA \)), so \( DC = 10 + 10 = 20 \). Now check \( DB^2 + BC^2 = (10\sqrt{3})^2 + 10^2 = 300 + 100 = 400 \), and \( DC^2 = 20^2 = 400 \). So by Pythagoras, △DBC is right-angled ( \( DB^2 + BC^2 = DC^2 \) ), so △DBC is right triangle. Also, \( \angle D = m\angle DBA \) (from step1) is true, and △DBC is right triangle is true. The statement \( DB = DC \) is false ( \( DB = 10\sqrt{3} \approx 17.32 \), \( DC = 20 \) ), and △ABC is not obtuse.

Answer:

The two true statements are:

• \( \angle D = m\angle DBA \)
• \( \triangle DBC \) is a right triangle