Question

2 circles touch each other externally at p . ab is a common tangent to the circle touch in them at a and
b. find the value of angle apb

Answer 1

Consider the attached diagram,
AA' ⊥ AB & BB' ⊥ AB ⇒ AA' ║ BB'

let ∠BB'P = 2θ ⇒ ∠AA'P = 180°-2θ
∵ PB' = BB' ⇒ Δ BB'P is isosceles triangle
∴ ∠BPB' = ∠PBB' = 90°- θ
∵ PA' = AA' ⇒ Δ AA'P is isosceles triangle
∴ ∠APA' = ∠PAA' = θ
 ∠APB + ∠APA' + ∠BPB' = 180°
∠APB = 180° - (∠APA' + ∠BPB') = 180°- (θ + 90°- θ) = 90°

Answer 2

Step-by-step explanation:

consider the attached figure,

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinx

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135°

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅ AD/sin30°=DC/sin15°

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅ AD/sin30°=DC/sin15°AD/2=ODcos∅

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅ AD/sin30°=DC/sin15°AD/2=ODcos∅DC/2=ODcos(135-∅)

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅ AD/sin30°=DC/sin15°AD/2=ODcos∅DC/2=ODcos(135-∅)solving all 3....

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅ AD/sin30°=DC/sin15°AD/2=ODcos∅DC/2=ODcos(135-∅)solving all 3....tan∅=√3

consider the attached figure,from ΔABD, AB/BD=sin(15+x)/sinx (sine rule)from ΔABC, AB/BC=sin(x)/sin(15+x) (sine rule)hence, 2sin²x=sin²(15+x)(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinxcotx=(((+/-)√2)-cos15)/sin15 hence ∠ADC= 135° hence obtuse let ∠ADO=∅ AD/sin30°=DC/sin15°AD/2=ODcos∅DC/2=ODcos(135-∅)solving all 3....tan∅=√3∅=6