A chord pq of length 6a is drawn in a circle of radius 10a .The tangents to the circle at p and q meet at r.find the are enclosed by pr,qr and the minor arc pq
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I think I can do this for you but it would be very difficult to explain it without using a diagram which I cannnot do in this limited system.
However I will try to explain it. If you do not understand my attempt, just click on my profile and email me to discuss further.
Draw a circle centre O.
Draw a chord pq.
Draw the two tangents to the circle at p and q. Let these meet at r.
Join pO and qO. Join rO. Let rO meet pq at X. They will be at right-angles to each other.
pX = qX = 3a
Consider the right-angle triangle OXq.
sin qOX = 3a/10a
sin qOX = 3/10
qOx = 17.46°
Now consider triangle rqO which is right-angled at q.
We know Oq = 10a and angle qOr = 17.46°.
This means we can use right-angled trig to work out Or and rq in terms of a.
Can you see how to proceed from here ?
Let me know if you can't, but it will be some 8 hours or so before I can get back to you.
However I will try to explain it. If you do not understand my attempt, just click on my profile and email me to discuss further.
Draw a circle centre O.
Draw a chord pq.
Draw the two tangents to the circle at p and q. Let these meet at r.
Join pO and qO. Join rO. Let rO meet pq at X. They will be at right-angles to each other.
pX = qX = 3a
Consider the right-angle triangle OXq.
sin qOX = 3a/10a
sin qOX = 3/10
qOx = 17.46°
Now consider triangle rqO which is right-angled at q.
We know Oq = 10a and angle qOr = 17.46°.
This means we can use right-angled trig to work out Or and rq in terms of a.
Can you see how to proceed from here ?
Let me know if you can't, but it will be some 8 hours or so before I can get back to you.
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Denote the centre of the circle by O and the angle POQ by Θ.
From triangle OPQ split down the middle you get sin(½Θ)=3/10
and from this you can get tan(½Θ)=3/√(91).
From triangle OPR PR=10atan(½Θ)=30a/√(91)
Area triangle OPR=½(OP)(PR)=150a²/√(91)
so area OPRQ = 300a²/√(91).
Area sector OPQ = ½(10a)²Θ=50a²Θ
Θ in radians = 2sin^-1(3/10)= 0.6094385
Area sector OPQ= 30.469a²
So required area =(300/√91 - 30.469)a²
=0.98a² (2SF)
If you are unfamiliar with radians, and Θ is in degrees, then the
area of the sector is π(10a)²(Θ/360) and you would get the same answer.
From triangle OPQ split down the middle you get sin(½Θ)=3/10
and from this you can get tan(½Θ)=3/√(91).
From triangle OPR PR=10atan(½Θ)=30a/√(91)
Area triangle OPR=½(OP)(PR)=150a²/√(91)
so area OPRQ = 300a²/√(91).
Area sector OPQ = ½(10a)²Θ=50a²Θ
Θ in radians = 2sin^-1(3/10)= 0.6094385
Area sector OPQ= 30.469a²
So required area =(300/√91 - 30.469)a²
=0.98a² (2SF)
If you are unfamiliar with radians, and Θ is in degrees, then the
area of the sector is π(10a)²(Θ/360) and you would get the same answer.