Math, asked by nishalakra224, 1 year ago

Ab, ac are two radii of circle inclined at an angle of 60°. Upon ac a point p is taken such that a circle can be described with centre p to touch the first circle internally and also to touch the circle with ab as diameter, externally. If ab = 2 cm, then the length of ap is :-

Answers

Answered by Shaizakincsem
5

Thank you for asking this question, here is your answer:


M is the mid-point of AB, AM = BM = R/2

The two circles will touch together so it would be PM = R/2 + r

Radius of circle will be r

AP = AC - PC = R - r

In Triangle APM

cosA=AP²+AM²−PM²/ 2x AP x PM

cos(60°) = (R−r)²+(R/2)²−(R/2+r)² /2(R−r)xR/2

1/2 = R² + r² - 2Rr+R²/4−R²/4−r2−Rr/(R(R−r)

R(R−r) =2(−3Rr+R²)

R²−Rr=−6Rr+2R²

R−r=−6r+2R  

R=5r

r=R/5

AP =R−r

=R−R/5

=4R/5

If there is any confusion please leave a comment below.

Answered by keshavjindal141103
0

M is the mid-point of AB, AM = BM = R/2

The two circles will touch together so it would be PM = R/2 + r

Radius of circle will be r

AP = AC - PC = R - r

In Triangle APM

cosA=AP²+AM²−PM²/ 2x AP x PM

cos(60°) = (R−r)²+(R/2)²−(R/2+r)² /2(R−r)xR/2

1/2 = R² + r² - 2Rr+R²/4−R²/4−r2−Rr/(R(R−r)

R(R−r) =2(−3Rr+R²)

R²−Rr=−6Rr+2R²

R−r=−6r+2R  

R=5r

r=R/5

AP =R−r

=R−R/5

=4R/5

Similar questions