Dear student, ∠B =60° and ∠C= 40°In ∆ABC,∠A+∠B+∠C =180°⇒∠A+60°+40° =180°⇒∠A=180°-100°⇒∠A=80°Now AD is bisector of ∠A Now, ∠CAD=∠DAB⇒∠CAD=40°Now In ∆ACL,90°+40°+∠CAL=180°⇒∠CAL=50°So, ∠LAD=∠CAL-∠CAD⇒∠LAD=50°-40°⇒∠LAD=10° Regards

Please help in Q15

Dear student,

∠ B = 60 ° a n d ∠ C = 40 ° I n ∆ A B C, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ A + 60 ° + 40 ° = 180 ° \Rightarrow ∠ A = 180 ° - 100 ° \Rightarrow ∠ A = 80 ° N o w A D i s b i s e c t o r o f ∠ A N o w, ∠ C A D = ∠ D A B \Rightarrow ∠ C A D = 40 ° N o w I n ∆ A C L, 90 ° + 40 ° + ∠ C A L = 180 ° \Rightarrow ∠ C A L = 50 ° S o, ∠ L A D = ∠ C A L - ∠ C A D \Rightarrow ∠ L A D = 50 ° - 40 ° \Rightarrow ∠ L A D = 10 °

Regards

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