Question

${\Large{\bf{Question :}}}$
1 :- Derivation of formula of Area of segment of Circle.

2 :- There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are the points directly opposite to each other on the two banks, and in a line with the tree. If the angles of elevation of the top of tree from P and Q are 30° and 45° respectively, find the height of the tree.

❎Dont Spam
❎Don't Copy from Google
✅Explain step by step
❎On the spot Spammed Answer will be reported
​

Answer 1

$\sf \Large (1)$

$\sf \large Theorem:$

$\sf Let\;C\;be\;a\;circle\;with\;radius=r$

$\sf Let\;S=segment\;of\;a\;circle\;C\;such\;that\;the\;base\;of\;the\;circle\;subtends\;an\;angle\;at\;the\;centre\;of\;the\;circle.$

$\sf Then,\;the\;(A)\;area\;of\;segment\;of\;circle\;(S)\;is\;given\;by:$

$\sf \boxed{\bigstar A=(\frac{1}{2})\times r^{2}(\theta -\sin \theta)}$

$\sf \to Here,\;\theta\;is\;measured\;in\;radians.$

$\sf \large Proof:$

$\sf Here,\;let\;BDCE\;be\;the\;segment\;S.$

$\sf Let\;b\;be\;the\;length\;of\;the\;base\;of\;S.$

$\sf Let\;BACE\;be\;the\;sector\;of\;C\;whose\;angle\;is\;\theta.$

$\sf The\;A\;is\;equal\;to\;the\;area\;of\;BACE-the\;area\;of\;the\;isosceles\;triangle\;\Delta ABC\;whose\;base\;is\;b.$

$\sf Let\;h\;be\;the\;altitude\;of\;\Delta ABC.$

$\sf \to From\;Area\;of\;Sector,\;we\;know\;that\;Area\;of\;Sector\;BACE\;in\;radians=(\frac{1}{2})\times r^{2}\theta$

$\sf \to From\;Area\;of\;Isosceles\;Triangle,\;the\;Area\;of\;\Delta ABC=(\frac{1}{2})\times r^{2}\sin\theta$

$\sf \to Thus, \;Area\;of\;Segment\;of\;circle,\;A=(\frac{1}{2})\times r^{2}(\theta -\sin\theta)$

$\sf \large Formulas\;to\;remember!!$

$\sf \to Area\;of\;a\;Segment\;of\;Circle\;in\;Radians=(\frac{1}{2})\times r^{2}(\theta -\sin\theta)$

$\sf \to Area\;of\;a\;Segment\;of\;Circle\;in\;Degrees=(\frac{1}{2})\times r^{2}[(\frac{\pi}{180})\theta -\sin\theta]$

$\sf \Large (2)$

$\sf \large Given,$

$\sf \to River=100m\;wide$

$\sf \to Angle\;of\;elevation\;of\;the\;top\;of\;tree\;from\;P=30^\circ$

$\sf \to Angle\;of\;elevation\;of\;the\;top\;of\;tree\;from\;Q=45^\circ$

$\sf \large To\;find,$

$\sf \to The\;height\;of\;the\;tree.$

$\sf \large Solution:$

$\sf Let\;OA\;be\;the\;tree\;of\;height\;h\;metre.$

$\sf In\Delta BOA\;and\;\Delta COA,\;\tan 30^\circ=\frac{OA}{OB}\;and\;\tan 45^\circ=\frac{OA}{OC}$

$\sf \to \frac{1}{\sqrt{3}}=\frac{h}{OB}\;and\;1=\frac{h}{OC}$

$\sf \to OB=\sqrt{3}h\;and\;OC=h$

$\sf \to OB+OC=\sqrt{3}h+h$

$\sf \to BC=(\sqrt{3}+1)h$

$\sf \to 100=(\sqrt{3}+1)h\quad[\because\;BC=100m]$

$\sf \to h=\frac{100}{\sqrt{3}+1}m$

$\sf \to h=\frac{100(\sqrt{3}+1)}{2}m$

$\sf \to h=50(1.73-1)m=36.6m$

$\sf \bigstar The\;height\;of\;the\;tree=36.6m$

Answer 2

1) If you know the radius, r, of the circle and you know the central angle, ϴ, in degrees of the sector that contains the segment, you can use this formula to calculate the area, A, of only the segment: A = ½ × r^2 × ((π/180) ϴ - sin ϴ) .

2) ANSWER refer to the image.

hope it helps u lot