Mathematics, 19.09.2019 17:50 ieshaking28

Ablimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the west of the blimp. if those measurements are 75.2° and 17.9°, how far apart are the two stadiums?

Ablimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the w

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Answer from: sonyfan

3115.03301179982 meters

This value is approximate

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Refer to the image attachment

Segments
AB = 1100 meters = height of blimp
DE = x meters = distance between stadiums
AD = y meters
AE = AD+DE = y+x

Angles
Angle ADB = angle CBD = 75.2 degrees
Angle CBE = angle BEA = 17.9 degrees

Focus on triangle ADB
tan(angle) = opposite/adjacent
tan(D) = AB/AD
tan(75.2) = 1100/y
y*tan(75.2) = 1100
y = 1100/tan(75.2)
y = 290.632536450073

Focus on triangle AEB
tan(angle) = opposite/adjacent
tan(E) = AB/AE
tan(E) = 1100/(y+x)
tan(17.9) = 1100/(290.632536450073+x)
0.32299119934583 = 1100/(290.632536450073+x)
0.32299119934583(290.632536450073+x) = 1100
93.8717515169298+0.32299119934583x = 1100
0.32299119934583x = 1100-93.8717515169298
0.32299119934583x = 1006.12824848308
x = 1006.12824848308/0.32299119934583
x = 3115.03301179982

This answer is approximate

Ablimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the w

Ablimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the w

Answer from: chrischris1

3315.03 meters.

Step-by-step explanation:

We have been given a blimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the west of the blimp. We are asked to find the distance between both stadiums.

First of all, we will find the distance between 1st stadium and base of blimp. We can see that stadium and blimp forms a right triangle with respect to ground, where m is adjacent side and 1100 meters is opposite side for 75.2 degrees angle. So we can set an equation as:

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

$\text{tan}(75.2^{\circ})=\frac{1100}{m}$

$m=\frac{1100}{\text{tan}(75.2^{\circ})}$

$m=\frac{1100}{3.784848088366}$

m=290.632536

Similarly, we will find value of x as:

$\text{tan}(17.9^{\circ})=\frac{1100}{m+x}$

$\text{tan}(17.9^{\circ})=\frac{1100}{290.632536+x}$

$290.632536+x=\frac{1100}{\text{tan}(17.9^{\circ})}$

$290.632536+x=\frac{1100}{0.322991199346}$

290.632536+x=3405.6655482

290.632536-290.632536+x=3405.6655482-290.632536

x=3115.033

$x\approx 3115.03$

Therefore, the two stadiums are approximately 3115.03 meters apart.

Answer from: Quest

honeycrisps have skyrocketed to about $4.50 per pound

Answer from: Quest

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .

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