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odolwa99
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Having a fair bit of trouble with this one. Based on the answer I had thought to apply the Sine Rule but adding points to describe the lengths of elevation seems wrong, but I'm stuck on how else to approach this. Can anyone help, please?
Many thanks.
Q. A vertical flagpole stands on horizontal ground. The angle of elevation of the top of the pole from a certain point on the ground is [itex]\theta[/itex]. From a point on the ground 10m closer to the pole, the angle of elevation is [itex]\beta[/itex]. Show that the height of the pole is [itex]\frac{10sin\theta sin\beta}{sin(\beta-\theta)}[/itex].
Points of pole: top c, bottom d.
Point at elevation [itex]\theta[/itex] = a
Point at elevation [itex]\beta[/itex] = b
[itex]sin\theta=\frac{a}{sin\theta}{|ac|}[/itex] = [itex]|ac|=\frac{a}{sin\theta}[/itex]
[itex]sin\beta=\frac{b}{|bc|}[/itex] = [itex]|bc|=\frac{b}{sin\beta}[/itex]
[itex]|\angle acb|180^o-(sin\theta+90^o) = 180^o-sin\theta[/itex]
I'd keep going at this point but I have a feeling I'm way off...
Many thanks.
Homework Statement
Q. A vertical flagpole stands on horizontal ground. The angle of elevation of the top of the pole from a certain point on the ground is [itex]\theta[/itex]. From a point on the ground 10m closer to the pole, the angle of elevation is [itex]\beta[/itex]. Show that the height of the pole is [itex]\frac{10sin\theta sin\beta}{sin(\beta-\theta)}[/itex].
The Attempt at a Solution
Points of pole: top c, bottom d.
Point at elevation [itex]\theta[/itex] = a
Point at elevation [itex]\beta[/itex] = b
[itex]sin\theta=\frac{a}{sin\theta}{|ac|}[/itex] = [itex]|ac|=\frac{a}{sin\theta}[/itex]
[itex]sin\beta=\frac{b}{|bc|}[/itex] = [itex]|bc|=\frac{b}{sin\beta}[/itex]
[itex]|\angle acb|180^o-(sin\theta+90^o) = 180^o-sin\theta[/itex]
I'd keep going at this point but I have a feeling I'm way off...