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3 years ago

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A surveyor is standing 50 feet from the base of a building. The angle of elevation to the top of the building is 60°. Approximat

ely how tall is the building?
25 feet

1 answer:

soldier1979 [14.2K]3 years ago

4 0

multiply the base ( 50 ) by tan(60)

50 x tan(60) = 86.6025

round answer as needed

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2 years ago

I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

3 years ago