1. Two straight roads intersect to form an angle of 75". Find the shortest distance from one road to a gas station
on the other road 10o0 m from the junction.
2. Two buildings with flat roofs are 60 m apart. From the roof of the shorter building, 40 m in height, the angle of
elevation to the edge of the roof of the taller building is 40". How high is the taller building?
3.A ladder, with its foot in the street, makes an angle of 30" with the street when its top rests on a building on
one side of the street and makes an angle of 40" with the street when its top rests on a building on the other side
of the street. I f the ladder is 50 ft long, how wide is the street'?
thanks for the answer..it will help a lot!!!!!!
1)
Draw a diagram. The shortest distance is the perpendicular from the road to the gas station
d = 100 sin75° = 96.59 m
2)
The height of the taller building above the shorter is
h = 60 tan40° = 50.34 m
3)
The distance to the shorter building is
50 cos30° = 43.3 m
The distance to the taller building is 50 cos40° = 38.3 m
Note that there is still osme figuring to do on #2 and #3.
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1. Well, if you're looking for the shortest distance to the gas station, I suggest taking a helicopter. No traffic, no stops, just direct to the station in no time! Who needs roads anyway?
2. Ah, the classic case of trying to reach new heights. If the shorter building is feeling a little down about its height, just remind it that it's closer to the ground, which means less chance of falling! As for the taller building, well it's just reaching for the stars! Or maybe just reaching for the roof of the shorter building, which is 40 meters higher.
3. Who needs width when you have length? But hey, I'm here to help. If the ladder is 50 ft long and it's making different angles with the street, then the street must be doing some yoga poses. In any case, let's call the width of the street "w". Using some trigonometry magic, we have w = 50 * (sin(40") - sin(30")). And voila, you've got yourself the width of the street! Just make sure to thank Pythagoras for his triangle wisdom.
1. To find the shortest distance from one road to a gas station on the other road, we can use trigonometry.
Step 1: Draw a diagram of the situation. Label the intersection where the two roads meet as point O. The gas station can be represented as point G.
Step 2: Draw a line segment from point G perpendicular to one of the roads. Let's call the point where this line intersects the road as point A.
Step 3: Connect point A to point O. This line segment represents the shortest distance from the road to the gas station.
Step 4: We now have a right triangle OAG, where OA is the shortest distance we are looking for, OG is the known distance from the intersection to the gas station (1000 m), and angle OAG is given as 75 degrees.
Step 5: Apply trigonometric functions to find OA. Since we have the opposite (OA) and the hypotenuse (OG), we can use the sine function.
sin(angle) = opposite/hypotenuse
sin(75) = OA/1000
Solve for OA:
OA = 1000 * sin(75)
Calculate the value to get the final answer.
2. To find the height of the taller building, we can use the concept of similar triangles.
Step 1: Draw a diagram of the situation. Label the shorter building as Building A and the taller building as Building B.
Step 2: Draw a horizontal line segment connecting the top edge of Building A to the top edge of Building B.
Step 3: We now have two right triangles - one with a height of 40 m and another with an unknown height (let's call it x) for Building B.
Step 4: Apply the concept of similar triangles. The ratio between the heights of the two triangles (40/x) is equal to the ratio between the distances of the horizontal line segment from the bases of the two buildings.
Step 5: Since the buildings are 60 m apart, the ratio of the horizontal distance to the heights should be the same.
60/40 = x/40
Solve for x:
x = (60 * 40) / 40
Calculate the value to get the final answer.
3. To find the width of the street, we can use trigonometry.
Step 1: Draw a diagram of the situation. Label the two buildings on either side of the street as Building A and Building B.
Step 2: Draw a line segment from the top of the ladder at Building A to the foot of the ladder on the street.
Step 3: Draw another line segment from the top of the ladder at Building B to the foot of the ladder on the street.
Step 4: We now have two right triangles - one with an angle of 30 degrees and another with an angle of 40 degrees. Both triangles share the same height (50 ft) for the ladder.
Step 5: Apply trigonometric functions to find the width of the street. We will use the tangent function for both triangles.
tan(angle) = opposite/adjacent
For the triangle with a 30-degree angle:
tan(30) = width/50
Solve for width:
width = 50 * tan(30)
For the triangle with a 40-degree angle:
tan(40) = width/50
Solve for width:
width = 50 * tan(40)
Calculate the values to get the final answer.
How about trying these yourself, I gave you a previous solution.
They look like straight-forward right-angled triangle trig problems.