What methods can you use to solve a triangle?

Answers

Answer 1

Law of Sines, Law of Sines, Pythagorean Theorem, Trigonometric Ratios, Heron's Formula .These methods can help you solve triangles and find missing side lengths, angles, or the area of the triangle.

To solve a triangle, you can use various methods depending on the given information. The methods include:

1. Law of Sines: This method involves using the ratio of the length of a side to the sine of its opposite angle.

2. Law of Cosines: This method allows you to find the length of a side or the measure of an angle by using the lengths of the other two sides.

3. Pythagorean Theorem: This method is applicable if you have a right triangle, where you can use the relationship between the lengths of the two shorter sides and the hypotenuse.

4. Trigonometric Ratios: If you know an angle and one side length, you can use sine, cosine, or tangent ratios to find the other side lengths.

5. Heron's Formula: This method allows you to find the area of a triangle when you know the lengths of all three sides.
These methods can help you solve triangles and find missing side lengths, angles, or the area of the triangle.

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Related Questions

Assume the following for this question. Lower and Upper specification limits for a service time are 3 minutes and 5 minutes, respectively with the nominal expected service time at 4 minutes. The observed mean service time is 4 minutes with a standard deviation of 0.2 minutes. The current control limits are set at 3.1 and 4.9 minutes respectively.

Answers

The observed mean service time falls within the current control limits. We can conclude that the process is stable, the service time is in control, and it meets the required specifications.


1. Calculate the process capability index (Cpk) using the formula: Cpk = min((USL - mean)/3σ, (mean - LSL)/3σ), where USL is the upper specification limit, LSL is the lower specification limit, mean is the observed mean service time, and σ is the standard deviation.
2. Plug in the values: USL = 5 minutes, LSL = 3 minutes, mean = 4 minutes, σ = 0.2 minutes.
3. Calculate Cpk: Cpk = min((5-4)/(3*0.2), (4-3)/(3*0.2)) = min(0.556, 0.556) = 0.556.
4. Since the calculated Cpk is greater than 1, the process is considered capable and the service time is in control.
5. The current control limits (3.1 and 4.9 minutes) are wider than the specification limits (3 and 5 minutes) and the observed mean (4 minutes) falls within these control limits.
6. Therefore, the process is stable and meets the specifications.

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Write six different iterated triple integrals for the volume of the tetrahedron cut from the first octant by the plane xyz. Evaluate the first integral. Question content area bottom Part 1

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Using triple integration, the volume of tetrahedron cut from the plane 2x + y + z = 4 is [tex]\frac{16}{3}[/tex].

A tetrahedron is nothing but a three dimensional pyramid.

To find the volume of tetrahedron cut from the plane 2x + y + z = 4, we need to first take one of the three dimension as base. Let as take xy plane as base.

XY as plane implies z = 0, equation becomes 2x + y = 4. To find the limits of X and Y, we put y = 0.

Thus, 2x + 0 = 4 , implying, x = 2.

Thus the range of x is : [0,2]

Putting the value of x in the given equation, the range of y is [0, 4 - 2x]

Similarly, range of z becomes: [0, 4 - 2x - y]

Since z is dependent upon y and x, and, y is dependent on x, Therefore the order of integration must be z, then y and then x.

The volume of tetrahedron becomes:

[tex]=\int\limits^0_2 \int\limits^{4-2x}_0 \int\limits^{4-2x-y}_0 {1} \, dz \, dy \, dx \\\\=\int\limits^0_2 \int\limits^{4-2x}_0 4-2x-y \, dy \, dx \\\\=\int\limits^0_2[ (4-2x)y - \frac{y^2}{2}]^{4-2x}_0 dx\\ \\=\int\limits^0_2 (4-2x)^2 - \frac{1}{2} (4-2x)^2 dx\\\\[/tex]

[tex]=\int\limits^2_0 {\frac{1}{2}(16+4x^2-16x )} \, dx \\\\=\int\limits^2_0(8+2x^2-8x)dx\\\\=[8x+\frac{2}{3} x^3-4x^2]^2_0\\\\=\frac{16}{3}[/tex]

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The complete question is given below:

Use triple integration to find the volume of tetrahedron cut from the plane 2x + y + z = 4.  

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