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
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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.
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