Respuesta :
a) Use the trigonometric function and ratio cos(x) = A/H. (Adjacent/Hypotenuse)
1. cos(x) = 135/H
2. H*cos(16°) = 135
3. H = 135/cos(16°)
***Make sure calculator is in degree mode***
4. H = 140.44 ft (rounded to two decimal places)
________
b) Use the function and ratio tan(x) = O/A (Opposite/Adjacent)
1. tan(x) = O/A
2. tan(16°) = O/135
3. 135*tan(16°) = O
4. O = 38.71 ft (two decimal places)
Good luck!
1. cos(x) = 135/H
2. H*cos(16°) = 135
3. H = 135/cos(16°)
***Make sure calculator is in degree mode***
4. H = 140.44 ft (rounded to two decimal places)
________
b) Use the function and ratio tan(x) = O/A (Opposite/Adjacent)
1. tan(x) = O/A
2. tan(16°) = O/135
3. 135*tan(16°) = O
4. O = 38.71 ft (two decimal places)
Good luck!
a) the distance of the man from is friend is 140.59 ft
b) the distance between his friend and the base of the cliff is 39.26 ft
How to calculate the distance of the man from is friend?
It is given that a man on a 135-ft vertical cliff looks down at an angle of 16°13' and sees his friend.
So, the vertical cliff, the line of sight of the man and the line joining his friend and the base of the cliff, forms a right angled triangle.
- The angle 16°13' is the angle between the the vertical cliff and the line of sight of the man.
- Here the distance of the man from is friend is actually the hypotenuse of the right angled triangle.
We know that, in a right angled triangle,
cosФ= [tex]\frac{base}{hypotenuse}[/tex], where Ф is the angle between the base and hypotenuse
In the right angle triangle we can write,
cos(16°13')= [tex]\frac{135}{hypotenuse}[/tex]
⇒ hypotenuse = 140.59 ft
∴ the distance of the man from is friend is 140.59 ft
How to find the distance from his friend and the base of the cliff?
Here, the distance between his friend and the base of the cliff is actually the perpendicular of the right angled triangle.
We know that, in a right angled triangle,
sinФ= [tex]\frac{perpendicular}{hypotenuse}[/tex], where where Ф is the angle between the base and hypotenuse
sin(16°13')= [tex]\frac{perpendicular}{140.59}[/tex]
⇒ perpendicular = 39.26 ft
∴ the distance between his friend and the base of the cliff is 39.26 ft
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