Question

f AC=BC, and BP ⃦CQ, then measure of x is​

Answer 1

Answer:

To Find:- value of x

Given:-

AC=BC

Solution:-

∠ACD+∠CDA+∠DAC=180°(sum of all interior angles of a triangle are supplementary)

∠ACD+x+x+14=180

∠ACD+2x+14=180 (Eq. 1)

In ∆ABC

∠ABC+∠BCA+∠CAB=180°(sum of all interior angles of a triangle are supplementary)

Given BC=AB

then

∠ABC=∠CBA (angles opposite to equal angles of a triangle are equal)

So,let them be x

that is x+x=2x (Eq. 2)

Also

∠CAB=32°(vertically opposite angles are equal)

therefore

So,from (Eq. 2)

∠CAB+2x=180

32+2x=180

2x=180-32

2x=148

x=148÷2

x=74°

Therefore In ∆ABC

∠CAB=32°

∠ABC=74°

∠BCA=74°

and now

In ∆ACD

∠ACD+∠BCA=180°(adjacent angles are supplementary)

∠ACD+74°=180

∠ACD=180-74°

∠ACD=106°

And now to find x

from (Eq. 1)

∠ACD+x+x+14=180

106+2x+14=180

2x+120=180

2x=180-120=60°

x=60/2

x=30°

Therefore the angles are

In ∆ACD

∠ACD=106°

∠CAD=x=30°

∠ADC=x+14=30+14=44°

In ∆ABC

∠BAC=32°

∠ABC=74°

∠BCA=74°

Hope it helps

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