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.
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.
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`
`Phat(A)R = Chat(R)Q`
Also you can further deduce considering BCP and ABQ triangles,
`Ahat(Q)B = Chat(P)B`
`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)`
AB = 100 and AQ = 120 and CP = 150.
`100/(BC) = (120)/(150)`
BC = 125
`120/150 = (BQ)/30`
`BQ = 24`
`BC = BQ + QC `
`125 = 24 + QC`
`QC = 125 -24 = 101`
Therefore the length of CQ is 101 cm
`` `\angle` APC = `\angle` ``AQC
Let `````\angle` APC = `\angle` AQC=`\theta`
AQ = 120
CP = 150
AP = 70, PB = 30 implies, AB = 100
`\angle` AQB = 180-`\angle` AQC = 180- `\theta`
`\angle` CPB = 180−`\angle` CPA = 180-`\theta`
`\angle`AQB = `` `\angle` CPB
`\angle` ``ABQ =`\angle` ``CBP
`\Delta` ABQ is similar to `\Delta` CBP by AA property.
By similar triangles:
CQ = BC-BQ = 125-24 =101
Therefore length CQ = 101
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
Yeah that's correct, nothing to worry about.
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