Join T C, viz. a Line drawn from the Tangent to the Center of the Circle, which Line will therefore represent the Semidiameter of the Earth, viz. 3958 Miles, according to Newton.
Draw a Line from C to M, which will pass throu’ some Point of the Circumference as H, the Base of the Mountain.
Then, in the Triangle M T C, as the Angle at T is a right Angle (Euclid’s Elements, Book 3, Proposition 18;) and the Sides M T, and T C, containing the right Angle, are known; the third Side C M is readily found: (being a Corollary to the 47th Prop. 1st Book Euclid:) viz. having the two Sides of a right Angle Triangle given to find the third. Therefore
RULE.
Multiply the Sides containing the right Angle, each into itself: viz. 150 and 3958: add the Products into one Sum: from which extract the square Root; equal to the Length in Miles, of the third Side required.
From the third Side, subtract that Part, viz. C H, which is equal to the Semidiameter T C already found: and the Remainder H M is the Height of the Mountain.
| Thus: | 150 | Miles. | 3958 | Miles in the Semidiameter of |
150 | 3958 | the Earth | ||
—— | ——— | |||
7500 | 31664 | |||
15 | 19790 | |||
——— | 35622 | |||
22500 | 11874 | |||
| Square of the | ———— | |||
| greatest visible | 15665764 | Square of the Semidiameter | ||
| Distance. | add 22500 | of the Earth. | ||
———— | ||||
| Extract the sq. Root, | 15688264 | (3960.84 Square Root. | ||
9 | 3958 subtract. | |||
— | —— | |||
69) 668 | Rem. 2.84 Answer in Miles. | |||
621 | ||||
—— | ||||
786) 478.2 | ||||
471 6 | ||||
——— | ||||
79208) 6664.00 | continued to 2 Decimals. | |||
6336 64 | ||||
———— | ||||
792164) 32736.00 | ditto. | |||
31686 56 | ||||
———— | ||||
104944 | ||||
To find the .84 Part of a Mile; multiply
| 1760 | Yards in a Mile, | |
| Decimal Parts of a Mile to be reduced | .84 | into Yards. |
| —— | ||
| 7040 | ||
| 14080 | ||
| ——— | ||
| 1760)1478.40 | (0 | |
| Subtract | 1478 | |
| —— | ||
| 282 |
Answer: the Height of the Mountain is 2 Miles 282 Yards.