ABC is a triangular park with AB =AC=100m .A clock tower is situated at midpoint of BC. The angles of elevation of top of tower at A and B are given by Cot alpha =3.2 and cosecent bita =2.6 respectively find height of tower
Answers
Answer:
25 m
Step-by-step solution:
cotα=3.2,cotβ=2.6
cotα=3.2,cotβ=2.6The triangle being isosceles, AB = AC the median AO is perpendicular to base BC
cotα=3.2,cotβ=2.6The triangle being isosceles, AB = AC the median AO is perpendicular to base BCOA=hcotα=3.2h
cotα=3.2,cotβ=2.6The triangle being isosceles, AB = AC the median AO is perpendicular to base BCOA=hcotα=3.2hOB=hcotβ=h
(cosec
(cosec 2
(cosec 2 β−1)
(cosec 2 β−1)
(cosec 2 β−1) =(2.4)h
(cosec 2 β−1) =(2.4)h100
(cosec 2 β−1) =(2.4)h100 2
(cosec 2 β−1) =(2.4)h100 2 =OA
(cosec 2 β−1) =(2.4)h100 2 =OA 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 100
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2 +6
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2 +6 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2 +6 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2 +6 2
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2 +6 2 =h.4.10
(cosec 2 β−1) =(2.4)h100 2 =OA 2 +OB 2 =h 2 [ 10032 2 +24 2 ]100×10=h.4 8 2 +6 2 =h.4.10∴h=25m
Step-by-step explanation:
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