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Rom4ik [11]
3 years ago
11

In ΔGHI, the measure of ∠I=90°, the measure of ∠G=82°, and GH = 3.4 feet. Find the length of HI to the nearest tenth of a foot.

Mathematics
2 answers:
taurus [48]3 years ago
5 0

Answer:

3.4

Step-by-step explanation:

Cloud [144]3 years ago
5 0

Answer:

3.4

Step-by-step explanation:

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From a point A that is 8.20 m above level ground, the angle of elevation of the top of a building s 31 deg 20 min and the angle
Mariulka [41]

Answer:

30.11 meters ( approx )

Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ 1\text{ minute } =\frac{1}{60}\text{ degree }

20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}

50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}

By the below diagram,

tan ( 12^{\circ} 50') = \frac{8.20}{y}

tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}

tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}

\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}

Now, again by the below diagram,

tan (31^{\circ}20')=\frac{x}{y}

tan(31+\frac{1}{3})=\frac{x}{y}

\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

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- hope this helps!!

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Answer:

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Step-by-step explanation:

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