Question

(163. AB is the diameter of the given circle and ZDAB = 40°, then find the value of
ZDBA
70​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{\mathcal{Solution}}}}}$

We know that for a semicircular triangle ,here

∆ABD,

$\rightarrow \angle ADB=90\degree$

Given

$\rightarrow \angle DAB=40\degree$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

now we know that sum of angles of s triangle is 180°

$\implies \angle DAB+\angle DBA+\angle ADB=180\degree\\</p><p>\implies 40\degree+\angle DBA+90\degree=180\degree\\ \implies \angle DBA=180\degree-(40\degree+90\degree)\\</p><p>\implies \angle DBA=180\degree-130\degree\\ \bf\implies \boxed{\red{\bf\angle DBA=50\degree}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \&#35;\mathcal{answer with quality }\: \: \&#38; \: \: \&#35;BAL }$