Question

NO LINKS!! URGENT HELP PLEASE!!!

Draw and label a diagram, then solve for the missing part. Round to the nearest tenth.
Please help me with #5 and 6

Answer 1

Answer 5 :-

To find:-

• Distance of swimmer from lifeguard.

Answer:-

We are here given that the line of sight is 8ft above the ground and the angle of depression is 18° . So ,

`\implies \angle DCA =\angle BAC = 18^o \\`

as these are alternate interior angles.

Now in ∆ABC ,

`\implies \tan \theta =(p)/(b)=(BC)/(AB) \\`

`\implies \tan 18^o =(8\ ft)/(AB) \\`

`\implies 0.325 = (8\ ft )/(AB)\\`

`\implies AB =(8)/(0.325)\ ft\\`

`\implies \underline{\underline{ AB = 24.61\ ft }}\\`

Hence the horizontal distance of the swimmer from the lifeguard is 24.61 ft.

Also , there is a second distance of the swimmer from the lifeguard that is along AC ,

`\implies \sin\theta =(BC)/(AC) \\`

`\implies \sin 18^o =(8\ ft )/(AC) \\`

`\implies AC = (8\ ft)/(\sin 18^o)=(8)/(0.309)\\`

`\implies \underline{\underline{ AC = 25.89 \ ft }} \\`

Hence the distance of the swimmer from the lifeguard along AC is 25.89ft .

`\rule{200}2`

Answer 6 :-

To find:-

• The altitude of the helicopter.

Answer :-

We are here given that the angle of depression is 73° . Therefore here ,

`\implies \angle SRP = \angle RPQ = 73^o \\`

as they are alternate interior angles.

Now in ∆PQR , we have;

`\implies \tan\theta = (p)/(b)=(QR)/(PQ) \\`

`\implies tan 73^o =(RQ)/(1200 \ ft) \\`

`\implies RQ = 1200\tan 73^o \ ft\\`

`\implies RQ = 1200 * 3.27 \ ft \\`

`\implies \underline{\underline{RQ = 3924 \ ft }} \\`

Hence the altitude of the helicopter is 3924 ft .

and we are done!