in question number 2 from the topic test-RHS congruence rule, they have proved that angle DBC=angleACB that is equal to - Maths - Congruence of Triangles

Question

in question number 2 from the topic test-RHS congruence rule,
they have proved that angle DBC=angleACB that is equal to 60degrees
so to find angle BDF we have to add anglesDBC and DCB that is
supposed to be 60 degrees+60 degrees, but in the soluions, they
have added 90degrees and 60degrees My question is why did they add
90degrees+60degrees when they actually had to add 60+60 and the
answer then will be 120 degrees, but the solutions shows the answer
is 150degress why did they add 90+60 when they had to add 60+60

Manbar Singh · Accepted Answer

Here is the complete solution

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In ∆ ABC and ∆DCB, ∠ABC = ∠DCB     90° each       AC = DB            given       BC = BC             common⇒∆ ABC ≅ ∆DCB  RHS⇒∠ACB = ∠DBC   CPCTNow, ∠ACB = 60°So,  ∠DBC = ∠ACB = 60°Since, ∠BDF is an exterior angle of ∆DCB, then ∠BDF = ∠DBC + ∠DCB   Exterior angle theorem⇒∠BDF = 60° + 90° = 150°

Hope this would have clear your doubt