Two ships are there in the Sea on either side of a lighthouse in such a way that the ship and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60degree &45degree. If the height of the light house is 200m , find the distances between two ships
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in triangle ABC
tan 60= ac/BC
√3=200/BC
√3bc=200. 1
in triangle ACD
tan 45 = ac/CD
1=200/CD
CD=200
bd =√3bc+200
bd=200√3+200
bd=200(√3+1)
tan 60= ac/BC
√3=200/BC
√3bc=200. 1
in triangle ACD
tan 45 = ac/CD
1=200/CD
CD=200
bd =√3bc+200
bd=200√3+200
bd=200(√3+1)
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