Question

The external bisectors of ∠B and ∠C of ΔABC met at point P. If ∠A = 100, ∠CPB = 8K, find K.

Answer 1

$\large\underline{\sf{Solution-}}$

It is given that,

• The external bisectors of ∠B and ∠C of ΔABC meet at point P.

• ∠A = 100

• ∠CPB = 8K

Now,

Since, BP bisects ∠ CBQ.

Let assume that, ∠QBP = ∠PBC = x say.

Also, .

It is given that, CP bisects ∠BCR.

Let assume that, ∠CRP = ∠PCB = y say.

Now,

Let further assume that, ∠ABC = a and ∠ACB = b.

Now,

Consider, Δ ABC,

We know, Sum of all interior angles of a triangle is supplementary.

So, 100° + a + b = 180°

$\bf\implies \:a + b = 80 \degree - - - (1)$

Now,

Consider, Δ BCP

Again, using Sum of all interior angles of a triangle is supplementary.

So,

$\rm :\longmapsto\:x + y + 8k = 180 \degree$

$\bf :\longmapsto\:x + y = 180 \degree - 8k - - - (2)$

Now,

Since, ABQ is a straight line.

$\rm :\longmapsto\:x + x + a = 180 \degree$

$\bf :\longmapsto\:2x + a = 180 \degree - - - (3)$

Again, As ACR is a straight line.

$\rm :\longmapsto\:y + y + b = 180 \degree$

$\bf :\longmapsto\:2y + b = 180 \degree - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:2y + b + 2x + a= 360 \degree$

$\rm :\longmapsto\:2(y + x)+ (b + a)= 360 \degree$

Now, using equation (1) and (2), we get

$\rm :\longmapsto\:2(180\degree \: - 8k)+ 80\degree= 360 \degree$

$\rm :\longmapsto\:360\degree \: - 16k+ 80\degree= 360 \degree$

$\rm :\longmapsto\: - 16k+ 80\degree= 0$

$\rm :\longmapsto\: - 16k = - 80\degree$

$\rm :\longmapsto\: k = 5\degree$

Hence,

• Value of k = 5°.

Short Cut trick,

1. If the external bisectors of ∠B and ∠C of ΔABC meet at point P, then ∠BPC = 90° - 1/2 ∠BAC.

2. If the internal bisectors of ∠B and ∠C of ΔABC meet at point P, then ∠BPC = 90° + 1/2 ∠BAC.

Answer 2

Answer:

\large\underline{\sf{Solution-}}

Solution−

It is given that,

The external bisectors of ∠B and ∠C of ΔABC meet at point P.

∠A = 100

∠CPB = 8K

Now,

Since, BP bisects ∠ CBQ.

Let assume that, ∠QBP = ∠PBC = x say.

Also, .

It is given that, CP bisects ∠BCR.

Let assume that, ∠CRP = ∠PCB = y say.

Now,

Let further assume that, ∠ABC = a and ∠ACB = b.

Now,

Consider, Δ ABC,

We know, Sum of all interior angles of a triangle is supplementary.

So, 100° + a + b = 180°

\bf\implies \:a + b = 80 \degree - - - (1)⟹a+b=80°−−−(1)

Now,

Consider, Δ BCP

Again, using Sum of all interior angles of a triangle is supplementary.

So,

\rm :\longmapsto\:x + y + 8k = 180 \degree:⟼x+y+8k=180°

\bf :\longmapsto\:x + y = 180 \degree - 8k - - - (2):⟼x+y=180°−8k−−−(2)

Now,

Since, ABQ is a straight line.

\rm :\longmapsto\:x + x + a = 180 \degree:⟼x+x+a=180°

\bf :\longmapsto\:2x + a = 180 \degree - - - (3):⟼2x+a=180°−−−(3)

Again, As ACR is a straight line.

\rm :\longmapsto\:y + y + b = 180 \degree:⟼y+y+b=180°

\bf :\longmapsto\:2y + b = 180 \degree - - - (4):⟼2y+b=180°−−−(4)

On adding equation (3) and (4), we get

\rm :\longmapsto\:2y + b + 2x + a= 360 \degree:⟼2y+b+2x+a=360°

\rm :\longmapsto\:2(y + x)+ (b + a)= 360 \degree:⟼2(y+x)+(b+a)=360°

Now, using equation (1) and (2), we get

\rm :\longmapsto\:2(180\degree \: - 8k)+ 80\degree= 360 \degree:⟼2(180°−8k)+80°=360°

\rm :\longmapsto\:360\degree \: - 16k+ 80\degree= 360 \degree:⟼360°−16k+80°=360°

\rm :\longmapsto\: - 16k+ 80\degree= 0:⟼−16k+80°=0

\rm :\longmapsto\: - 16k = - 80\degree:⟼−16k=−80°

\rm :\longmapsto\: k = 5\degree:⟼k=5°

Hence,

Value of k = 5°.

Short Cut trick,

1. If the external bisectors of ∠B and ∠C of ΔABC meet at point P, then ∠BPC = 90° - 1/2 ∠BAC.

2. If the internal bisectors of ∠B and ∠C of ΔABC meet at point P, then ∠BPC = 90° + 1/2 ∠BAC.