Use the cosine law to find the length of side c:
![\begin{gathered} c^2=1^2+9^2-2(1)(9)\cos 60 \\ c^2=1+81-18\cos 60 \\ c^2=82-18\cos 60 \\ c^2=82-9 \\ c^2=73 \\ c=\sqrt[]{73} \\ c\approx8.54 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20c%5E2%3D1%5E2%2B9%5E2-2%281%29%289%29%5Ccos%2060%20%5C%5C%20c%5E2%3D1%2B81-18%5Ccos%2060%20%5C%5C%20c%5E2%3D82-18%5Ccos%2060%20%5C%5C%20c%5E2%3D82-9%20%5C%5C%20c%5E2%3D73%20%5C%5C%20c%3D%5Csqrt%5B%5D%7B73%7D%20%5C%5C%20c%5Capprox8.54%20%5Cend%7Bgathered%7D)
Use the cosine law to find the measure of angles A or B:
![\begin{gathered} \\ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ \\ To\text{ find angle A:} \\ \frac{\sin A}{a}=\frac{\sin C}{c} \\ \\ \frac{\sin A}{1}=\frac{\sin 60}{\sqrt[]{73}} \\ \\ \sin A=\frac{\sin 60}{\sqrt[]{73}} \\ \\ A=\sin ^{-1}(\frac{\sin60}{\sqrt[]{73}}) \\ \\ A\approx5.82º \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%20%5C%5C%20%5Cfrac%7B%5Csin%20A%7D%7Ba%7D%3D%5Cfrac%7B%5Csin%20B%7D%7Bb%7D%3D%5Cfrac%7B%5Csin%20C%7D%7Bc%7D%20%5C%5C%20%20%5C%5C%20To%5Ctext%7B%20find%20angle%20A%3A%7D%20%5C%5C%20%5Cfrac%7B%5Csin%20A%7D%7Ba%7D%3D%5Cfrac%7B%5Csin%20C%7D%7Bc%7D%20%5C%5C%20%20%5C%5C%20%5Cfrac%7B%5Csin%20A%7D%7B1%7D%3D%5Cfrac%7B%5Csin%2060%7D%7B%5Csqrt%5B%5D%7B73%7D%7D%20%5C%5C%20%20%5C%5C%20%5Csin%20A%3D%5Cfrac%7B%5Csin%2060%7D%7B%5Csqrt%5B%5D%7B73%7D%7D%20%5C%5C%20%20%5C%5C%20A%3D%5Csin%20%5E%7B-1%7D%28%5Cfrac%7B%5Csin60%7D%7B%5Csqrt%5B%5D%7B73%7D%7D%29%20%5C%5C%20%20%5C%5C%20A%5Capprox5.82%C2%BA%20%5Cend%7Bgathered%7D)
Answer:
<em>y</em> = (5/3)<em>x</em> - 9
Step-by-step explanation:
First, find the slope (rise over run):
m = (-9 - (-14)) / (0 - (-3)) = (-9 + 14) / (0 + 3) = 5/3
Use the point-slope form equation with the given points (-3, -14):
y + 14 = (5/3)(x + 3)
or optionally in slope intercept form:
<em>y</em> = (5/3)<em>x</em> - 9
divide 18 by 24 for a decimal number:
18 / 24 = 0.75
multiply 0.75 by 100 for the percent
0.75 * 100 = 75%
Answer:
789 m²
Step-by-step explanation:
Consider the cross section created by a vertical plane through the apex of the pyramid and bisecting opposite sides. The cross section is an isosceles triangle with base 20 m and height 17 m. One side of this triangle is the slant height of the face of the pyramid.
The side of the triangle above can be found using the Pythagorean theorem. A median from the apex of the triangle will divide it into two right triangles, each with a base of 10 m and a height of 17 m. Then the hypotenuse is ...
s² = (10 m)² +(17 m)² = 389 m²
s = √389 m ≈ 19.723 m . . . . . slant height of one triangular face
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The area of one triangular face is ...
A = (1/2)sb
where s is the slant height above, and b is the 20 m base of the face of the pyramid. There are 4 of these faces, so the total area is ...
total lateral area = 4A = 4(1/2)sb = 2sb = 2(19.723 m)(20 m)
total lateral area ≈ 789 m²
Don’t know but you do have to subtract
