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Tuberculosis.
Answer:
ANSWER
Suppose the vertices of the sector are labelled A,B and C with angle BAC =θ.
Given diameter =14cm ⟹ Radius (R)=7cm
Then ∣AB∣=∣AC∣=R=7.
From A, draw the bisector m of θ in the sector.
This line m goes through the centre of the circumscribing circle and intersects that circle at point S(opposite to vertex A).
The length AS=2r, where r is the radius that we need to determine.
Draw from S the line segment SB.
Since AS goes through the centre of the circumscribing circle and B is a point on that same circle,
△ASB is a right triangle and ∠SBA=90o.
Furthermore, angle BAS=θ/2
So ∣AS∣=cos(θ/2)∣AB∣
Since ∣AS∣=2r and ∣AB∣=R=7, we have
2r=cos(θ/2)R=cos(θ/2)7
⟹ r=21×cos(θ/2)7=27sec(θ/2)
∴ Area of the circle circumscribing the sector ABC=πr2=722×472sec2(2
a sector with acute central angle theta is cut out fron a circle of radius 7cm area ofcircle circumscribing the sector is