what is the area of triangle PQR round to the nearest tenth of a square meter? drawing is not drawn to scale. 18m 25 degrees 8m
a) 60.8 m^2
b) 65.3 m^2
c) 30.4 m^2
d) 32.7 m^2
15 answers
Unfortunately, it's difficult to answer the question without seeing the drawing. Can you provide a diagram or additional information about triangle PQR?
Two side lengths are 18 and 8. One angle is 25 degrees
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To find the area of triangle PQR, we can use the formula:
Area = (1/2) x base x height
where the base is one of the sides and the height is the perpendicular distance from the base to the opposite vertex.
First, let's find the height of the triangle. We can use trigonometry, specifically the sine function:
sin(25) = opposite / 18
opposite = 18 x sin(25) ≈ 8.076
So the height of the triangle is approximately 8.076 meters.
Now we can use the formula:
Area = (1/2) x base x height
Area = (1/2) x 8 x 8.076 ≈ 32.6
Rounding to the nearest tenth of a square meter, the area of triangle PQR is approximately 32.7 m^2.
Therefore, the answer is d) 32.7 m^2.
Area = (1/2) x base x height
where the base is one of the sides and the height is the perpendicular distance from the base to the opposite vertex.
First, let's find the height of the triangle. We can use trigonometry, specifically the sine function:
sin(25) = opposite / 18
opposite = 18 x sin(25) ≈ 8.076
So the height of the triangle is approximately 8.076 meters.
Now we can use the formula:
Area = (1/2) x base x height
Area = (1/2) x 8 x 8.076 ≈ 32.6
Rounding to the nearest tenth of a square meter, the area of triangle PQR is approximately 32.7 m^2.
Therefore, the answer is d) 32.7 m^2.
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in triangle pqr p=53, pq=7.4, and pr=9.6. what is m r to the nearest degree?
To find angle P in triangle PQR, we can use the law of cosines:
c^2 = a^2 + b^2 - 2ab cos(C)
Where c is the side opposite the angle C. In this case, side PR is opposite angle P.
Substituting the given values:
9.6^2 = 7.4^2 + 53^2 - 2(7.4)(53) cos(P)
Solving for cos(P):
cos(P) = (7.4^2 + 53^2 - 9.6^2) / (2(7.4)(53))
cos(P) ≈ 0.886
Using the inverse cosine function, we can find the measure of angle P:
P ≈ cos^(-1)(0.886) ≈ 28.3 degrees
Therefore, the measure of angle R would be:
R ≈ 180 - 53 - 28.3 ≈ 98.7 ≈ 99 degrees (rounded to the nearest degree).
So, the measure of angle R is approximately 99 degrees.
c^2 = a^2 + b^2 - 2ab cos(C)
Where c is the side opposite the angle C. In this case, side PR is opposite angle P.
Substituting the given values:
9.6^2 = 7.4^2 + 53^2 - 2(7.4)(53) cos(P)
Solving for cos(P):
cos(P) = (7.4^2 + 53^2 - 9.6^2) / (2(7.4)(53))
cos(P) ≈ 0.886
Using the inverse cosine function, we can find the measure of angle P:
P ≈ cos^(-1)(0.886) ≈ 28.3 degrees
Therefore, the measure of angle R would be:
R ≈ 180 - 53 - 28.3 ≈ 98.7 ≈ 99 degrees (rounded to the nearest degree).
So, the measure of angle R is approximately 99 degrees.
in triangle pqr m p=53, pq=7.4, and pr=9.6. what is m r to the nearest degree?
In triangle PQR, the measure of angle R can be found using the law of cosines.
c^2 = a^2 + b^2 - 2ab cos(C)
where c is the side opposite to the angle C.
Substituting the given values:
9.6^2 = 53^2 + 7.4^2 - 2(53)(7.4) cos(R)
Solving for cos(R):
cos(R) = (53^2 + 7.4^2 - 9.6^2) / (2(53)(7.4))
cos(R) ≈ 0.947
Using the inverse cosine function, we get:
R ≈ cos^(-1)(0.947) ≈ 18.8 degrees
Therefore, the measure of angle R is approximately 18.8 degrees (rounded to the nearest degree).
c^2 = a^2 + b^2 - 2ab cos(C)
where c is the side opposite to the angle C.
Substituting the given values:
9.6^2 = 53^2 + 7.4^2 - 2(53)(7.4) cos(R)
Solving for cos(R):
cos(R) = (53^2 + 7.4^2 - 9.6^2) / (2(53)(7.4))
cos(R) ≈ 0.947
Using the inverse cosine function, we get:
R ≈ cos^(-1)(0.947) ≈ 18.8 degrees
Therefore, the measure of angle R is approximately 18.8 degrees (rounded to the nearest degree).