# In a triangle abc, p and q are points on the sides AB and CB respectively such that angle APC=angle AQC, AP=70cm, PB=30cm, AQ=120 cm and CP=150cm. Find the length of CQ.

justaguide | Certified Educator

In a triangle ABC, P and Q are points on the sides AB and CB respectively such that angle APC=angle AQC, AP=70cm, PB=30cm, AQ=120 cm and CP=150cm. The length of side CQ has to be determined.

Let the point of intersection of AQ and CP be R. Consider the triangles APR and CQR.

It is given that /_ APC = /_ AQC which also gives us that /_ APR = /_CQR .

The angles PRA and QRC are equal as they are vertical angles.

/_ PAR = 180 - (/_APR + /_PRA)

/_ QCR = 180 - (/_ AQC + /_CQR)

This gives /_PAR = /_QCR .

/_PAR and /_ BAQ refer to the same angle. So do /_QCR and /_BCP

In triangles BAQ and BCP

/_BAQ = /_BCP

The angle ABC is common.

Therefore, the triangles BAQ and BCP are similar triangles.

In similar triangles, corresponding sides are in the same proportion.

For triangles BAQ and BCP, we have,

(BA)/(BC) = (QA)/(PC) = (BQ)/(BP)

(AP+PB)/(BQ+QC) = (QA)/(PC) = (BQ)/(BP)

Using the length of sides provided,

(70+30)/(BQ+CQ) = 120/150 = (BQ)/30

(BQ)/30 = 120/150

=> (BQ) = 120*30/150 = 24

(70+30)/(BQ+CQ) = 120/150

=> (70+30)/(24+CQ) = 120/150

=> 100/(24+CQ) = 12/15

=> 24+CQ = 1500/12

=> CQ = 1500/12 - 24

=> CQ = 101

The length of side CQ is 101 cm.

thilina-g | Certified Educator

Let the point of intersection of CP and AQ be R. If you draw a diagram you would see that,

Consider APR and CQR traingles

Ahat(P)R = Chat(Q)R (Given)

Chat(R)Q = Ahat(R)P

This gives,

Phat(A)R = Chat(R)Q

Also you can further deduce considering BCP and ABQ triangles,

Ahat(Q)B = Chat(P)B

Also,

Phat(B)C = Chat(B)P   (Common angle)

This makes the two triangles, ABQ and BCP are similar in shape.

Therefore their respective sides has the same ratios.

(AB) / (BC) = (AQ)/(CP) = (BQ)/(BP)

We know,

AB = 100 and AQ = 120 and CP = 150.

100/(BC) = (120)/(150)

BC = 125

Also,

120/150 = (BQ)/30

BQ = 24

We know,

BC = BQ + QC

125 = 24 + QC

QC = 125 -24 = 101

Therefore the length of CQ is 101 cm

Neethu | Certified Educator

Given:
 \angle APC = \angle AQC

Let \angle APC = \angle AQC=\theta
AQ = 120
CP = 150
AP = 70, PB = 30 implies, AB = 100

Now,
\angle AQB = 180-\angle AQC = 180- \theta
\angle CPB = 180−\angle CPA = 180-\theta

\angleAQB =  \angle CPB
\angle ABQ =\angle CBP

\Delta ABQ is similar to \Delta CBP by AA property.

By similar triangles:
\frac{BC}{CP}=\frac{AB}{AQ}

\frac{BC}{150}=\frac{100}{120}
BC=\frac{100\times 150}{120}=125

\frac{BC}{BP}=\frac{AB}{BQ}

\frac{125}{30}=\frac{100}{BQ}

BQ=\frac{100\times 30}{125}=24

Now ,

CQ = BC-BQ = 125-24 =101

Therefore length CQ = 101

vaaruni | Student

Given : AP=70, BP=30, AQ= 120, PC=150 angle APC= angle AQC, Let the interecting points of the two lines AQ and CP be R. We take the two  triangles BPC and ABQ. The triangle BPC similar to triangle ABQ  by test A.A.A [ angle BPC(suplimentary to APC)= angle AQB(suplimentary to AQC), angle PBC=angle ABQ (common angles),Since two angles of one triangle = corresponding two angles of the another triangle therefore the remaining angle BCP=remaining angle BAQ  a                     Using properties (side proportionality)of the similar triangle we get,  AB/BC = AQ/PC = BQ/BP

100/BC = 120/150 [ taking AB/BC = AQ/PC and AB=AP+PB=70+30 ]

BC= (100*150)/120 = 125

Next, taking the ratio AB/BC = BQ/BP

100/125 = BQ/30

BQ=(100*30)/125 = 24

QC= BC - BQ = 125 - 24 = 101

zopalwie | Student

Yeah that's correct, nothing to worry about.

elekzy | Student

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