Question

In Fig (i) ,triangle ABC is right angled at A ,and AD perpendicular to BC .Also ,angle B =55°.Find (a)angle BAD​

Answer 1

In ∆ ABC,

$\tt∠A + ∠B + ∠C = 180° \: \: \: \tiny \tt\{ angle \: sum\: property \:of \: ∆\}$

Here,

$\huge \to \: \bf{}AD⊥BC$

=> ∠ADB = ∠ADC = 90°

Now,

In ∆ADB,

$\small \text{∠ADB + ∠ABD + ∠BAD = 180° } \: \: \: \tiny \tt\{ angle \: sum\: property \:of \: ∆\} \\ \\ \bf = > 90° + 55° + ∠BAD = 180° \\ \bf = > \: \: \: \: \: \: \: ∠BAD = 180° - (90 + 55)° \\ \bf = > \: \: \: \: \: \: \: ∠BAD = 180° - 145° \\ \large \boxed{\sf = > \: \: \: \: \: \: \: ∠BAD = \green{ 35°}}$