MHB Flagpole Angle: 40 Degrees - What's 3/4 of the Way Up?

[oldfart1]

a flagpole subtends a angle of 40 degrees.
What angle does it subten 3/4 of the way up.
Answer is supposedly 32 degrees ,but how ?

oldfart

 

[Greg]

Let $$y$$ be the height of the flagpole. Let $$x$$ be the distance from the base of the flagpole to the observer.
Then $$\tan(40^\circ)=\dfrac yx$$. Letting $$q$$ be the angle in question we have $$\tan(q)=\dfrac34\cdot\dfrac yx$$
Can you continue?

 

[oldfart1]

Let $$y$$ be the height of the flagpole. Let $$x$$ be the distance from the base of the flagpole to the observer.
Then $$\tan(40^\circ)=\dfrac yx$$. Letting $$q$$ be the angle in question we have $$\tan(q)=\dfrac34\cdot\dfrac yx$$
Can you continue?

Sorry but no. thanks

 

[SuperSonic4]

Sorry but no. thanks

As Greg pointed out you have $$\tan(q) = \dfrac{3}{4} \cdot \dfrac{y}{x}$$ together with $$\tan(40^{\circ}) = \dfrac{y}{x}$$

That means you can put in $$\tan(40^{\circ})$$ (from the second equation) in place of [math]\dfrac{y}{x}[/math] from the first.

This substitution gives you $$\tan(q) = \dfrac{3}{4} \cdot \tan(40^{\circ})$$

The right hand side of that equation is simply a number your calculator can find for you. Does that make it clearer?

 

[oldfart1]

Thanks for your help..I get it
Oldfart