Question

8. (i) In the given figure, AB = AD, angle PAD=70°
and angle DBC = 90°. Find x, angle BDQ and angle BCD
if AB II DC.
​

Answer 1

Step-by-step explanation:

$In\:\triangle ABD\\</p><p>AB = AD..... (Given) \\</p><p>\therefore \angle ABD = \angle ADB=x\\</p><p> \angle ABD + \angle ADB=70\degree\\... (By \:exterior \:angle \:theorem) \\</p><p>\therefore x + x=70\degree\\ </p><p>\therefore 2x=70\degree\\</p><p>\implies x =\frac{70\degree}{2}\\</p><p>\huge \fbox{\therefore x =35\degree} \\</p><p>\implies \angle ABD = \angle ADB =35\degree\\</p><p>\because \angle ADC = \angle PAD\\...(Alternate\: angles) \\</p><p>\therefore \angle ADB+ \angle BDC =70\degree\\</p><p>\therefore 35\degree + \angle BDC =70\degree\\</p><p>\therefore \angle BDC =70\degree-35\degree\\</p><p>\therefore \angle BDC = 35\degree\\</p><p>In\: \triangle BCD\\</p><p>\angle BDC +\angle BCD +\angle CBD = 180\degree \\</p><p>\therefore 35\degree+\angle BCD +90\degree=180\degree \\</p><p>\implies \angle BCD= 180\degree-125\degree\\</p><p>\huge\fbox{\therefore \angle BCD= 55\degree}\\</p><p>\angle BDQ = \angle ADB+\angle ADQ\\ \therefore \angle BDQ = 35\degree + (180\degree - 70\degree)\\</p><p>\therefore \angle BDQ = 35\degree + 110\degree \\</p><p>\huge\fbox{\therefore \angle BDQ = 145\degree}$