Q12 of 153 Page 166

In the given figure, BE ⊥ AC, ∠DAC = 30° and ∠DBE = 40°. Find ∠ACB and ∠ADB.

In triangle BEC we have,


B = 40O and E = 90O


So,C = 180O – (90 + 40)


=50O


Therefore,ACB = 50O


Now intriangle ADC we have,


A = 30O and C = 50O


So,D = 180O – (30 + 50)


=100O


Therefore,


ADB + ADC = 180 (sum of angles on straight line)


ADB + 100 = 180


ADB = 180 – 100


= 80O


More from this chapter

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10

In the given figure, AB || CD, ∠BAD = 30° and ∠ECD = 50°. Find ∠CED.

 11

In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.



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In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.

 14

In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.