Awoman measures the angle of elevation of a mountaintop as 12.0°. after walking 1.00 km closer to the mountain on level ground, she fi nds the angle to be 14.0°. (a) draw a picture of the problem, neglecting the height of the woman's eyes above the ground. hint: use two triangles. (b) select variable names for the mountain height (suggestion: y) and the woman’s original distance from the mountain (suggestion: x) and label the picture. (c) using the labeled picture and the tangent function, write two trigonometric equations relating the two selected variables. (d) find the height y of the mountain by fi rst solving one equation for x and substituting the result into the other equation.

The height of the mountain is 1.33 km.

Explanation:

a and b) See the attached figure for a description of the situation

c) For right triangles, this trigonometric rule applies:

tangent angle = opposite side / adjacent side

Then, in our problem:

tg 12º = y / x

tg 14º = y / (x-1)

d) using the first equation:

tg 12º = y / x

x * tg 12º = y

Replacing "y" for "x * tg 12º" in the second equation and solving for x:

tg 14º = (x * tg 12º) / (x-1)=

tg 14º * (x-1) = tg 12º * x

0.25 x -0.25 = 0.21 x

0.25 x - 0.21 x = 0.25

0.04 x = 0.25

x = 6.25

Then:

y = tg 12º * x

y = 6.25 * tg 12º = 1.33 km

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j