Finding an elevation of a building.

[zschnopz]

So this is on my homework, I require your assistance. Please. I've been working on it for many hours but I can't seem to crack the mysterious code.

The angle of elevation to the top of building H, from a point A, in building L, that is across the street from building H, and 38 feet above the ground, is 27 degrees. From point B, that is 21 feet vertically above point A, the angle of elevation to the top of the building H is 22 degrees. How tall is building H?

 

[stapel]

I've been working on it for many hours....

Please reply with a sample of your efforts so far, so we can see where you're getting bogged down.

The angle of elevation to the top of building H, from a point A, in building L, that is across the street from building H, and 38 feet above the ground, is 27 degrees. From point B, that is 21 feet vertically above point A, the angle of elevation to the top of the building H is 22 degrees. How tall is building H?

You've drawn two vertical lines, being the sides of the buildings. You've drawn a horizontal line between them, being the ground. You've drawn another horizontal line somewhere around the middle of the buildings, being the height of the viewer at A in Building L. You've drawn another horizontal line between the buildings, above the previous line, being the height of the viewer at B in Building L. You've labelled the distance between the buildings with some variable; I'll use "d". You've drawn slanty lines from A and B to the top of Building H, and labelled the two angles of elevation for the two right triangles you've formed. You've labelled the portion of the height on Building H between A's line and B's line as 21. You've labelled the rest of the height of Building H (from B's line to the top of Building H) with some variable; I'll use "h". You've created the two trig ratios. And... then what?

 

[zschnopz]

So this is what I have so far. So to solve for H you have to solve for D, right? and to solve for D you need the length of the hypotenuse from A(with 27 degrees) to the top of building H. I'm not really sure how to approach finding that. We're not supposed to use law of sines or anything which would make more sense. This is probably way more straight forward than I think it is. Thank you for the help
View attachment 3319

 

[DrPhil]

So this is what I have so far. So to solve for H you have to solve for D, right? and to solve for D you need the length of the hypotenuse from A(with 27 degrees) to the top of building H. I'm not really sure how to approach finding that. We're not supposed to use law of sines or anything which would make more sense. This is probably way more straight forward than I think it is. Thank you for the help
View attachment 3319

The attachment was invalid, so I don't see your two trig ratios. You must have two equations in the two unknowns, D and H. You don't have to solve for D explicitly - you can eliminate it by substitution and thus form a single equation with H the only unknown.

 

[zschnopz]

yeah, it does seem that there are useless values. haha. So if I tried substituting could I solve for the unknown length by using...tan (27) = (21+x)/d and tan (22) = x/d so that tan (27) = (21+(tan (22) * d)/d = d = 199.05. so then using tan (63) = 199.05/h = 80.39 + 21 + 38 = 139.39!

Is that right?

 

[DrPhil]

yeah, it does seem that there are useless values. haha. So if I tried substituting could I solve for the unknown length by using...tan (27) = (21+x)/d and tan (22) = x/d so that tan (27) = (21+(tan (22) * d)/d = d = 199.05. so then using tan (63) = 199.05/h = 80.39 + 21 + 38 = 139.39!

Is that right?

Yes