Example Question - solving right triangle
Here are examples of questions we've helped users solve.
Solving Right Triangle with Pythagorean Theorem
这个问题是一个直角三角形的问题，我们要使用勾股定理来求解斜边AB的长度。勾股定理表示在直角三角形中，斜边的平方等于两个直角边的平方和。所以，我们可以表示为： AB² = AC² + BC² 已知AC = 2米，BC = 6米。我们可以把它们代入上面的公式： AB² = 2² + 6² AB² = 4 + 36 AB² = 40 然后，我们需要计算AB的值，即求平方根： AB = √40 AB = 2√10 (约为6.32米) 因此，直角三角形斜边AB的长度约为6.32米。
Solving for Missing Leg Length of a Right Triangle
The given image shows a right triangle with one leg measuring 13 cm, and the hypotenuse measuring 23 cm. The question likely asks for the length of the other leg of the triangle. We can use the Pythagorean theorem to solve for the missing length. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (the legs). The formula can be written as: \[ c^2 = a^2 + b^2 \] where \( c \) is the length of the hypotenuse, and \( a \) and \( b \) are the lengths of the other two sides. Using this theorem and substituting the known values, we get: \[ 23^2 = 13^2 + b^2 \] \[ 529 = 169 + b^2 \] \[ b^2 = 529 - 169 \] \[ b^2 = 360 \] \[ b = \sqrt{360} \] \[ b = 19 \] So the length of the missing leg \( b \) is 19 cm.
Solving for Angle x in a Right Triangle
The problem in the image shows a right triangle (as indicated by the small square which denotes a 90-degree angle). You are asked to find the value of x, which is the measure of the angle opposed to the side of length 14. To solve for the angle x, you can use the trigonometric function tangent (tan), because it relates the opposite side to the adjacent side in a right triangle. \[ \tan(x) = \frac{opposite}{adjacent} \] \[ \tan(x) = \frac{14}{18} \] Now you'll use the arctangent function to find the angle whose tangent is 14/18. \[ x = \arctan\left(\frac{14}{18}\right) \] Using a calculator: \[ x \approx \arctan(0.7778) \] \[ x \approx 37.87 \text{ degrees} \] Rounded to the nearest degree, x is approximately 38 degrees.
Solving Right Triangle GHK
The image shows two right triangles: Triangle GHK (which is triangle number 4) and another triangle next to it (triangle number 5), for which the details are not visible. In triangle GHK, we have to solve for variables x, y, and z. Triangle GHK is a right triangle, with the right angle at H. Given: - Segment GH = 5 (which is the base of the triangle). - Segment HK = 15 (which is the hypotenuse of the triangle). - Segment GH is perpendicular to segment HK, forming the right angle at H. - Segment GJ = x (this segment is not visible in the provided image, but since J is not defined, this could be a typo and might refer to GH). - Segment HJ = z (this segment is not visible in the provided image, but since J is not defined, this could be a typo and might refer to HJ, the height). - Segment JK = y (which is the remaining side, opposite the right angle). To solve for x, y, and z, we will use the Pythagorean theorem for right triangles: a^2 + b^2 = c^2, where a and b are the legs of the triangle, and c is the hypotenuse. Let's assume z refers to the height of the triangle (segment HJ), x refers to the length of the hypotenuse (segment HK, already given as 15), and y refers to the length of the opposite side of the right angle, which is usually the hypotenuse but seems to be used differently in this context. The given length of the hypotenuse (HK) as 15 and the base (GH) as 5 doesn't require solving and thus, x = 15 from the given information. To solve for y and z, you would proceed as follows: Since GH is 5, HJ (height, which is z) completing the perpendicular side, can be calculated using the Pythagorean theorem with values 5 and 15: z^2 = 15^2 - 5^2 z^2 = 225 - 25 z^2 = 200 z = √200 z = √(100 * 2) z = 10√2 The length of z is 10√2. If y refers to the base opposite the height in this triangle, and with segment GH being part of segment GJ and the whole segment GJ (base) sums to 20 since it's the difference between HK (15) and HG (5); hence, segment JH will be: y = 15 - 5 = 10 So the three variables solved from this triangle are: x = 15 (the hypotenuse, given) y = 10 (the length of JK, assuming JK refers to the remainder of the base) z = 10√2 (the height of the triangle)
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