This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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Answer
225.61, mm
Here are the solutions to your questions:
2. Calculate the vertical length and both included angles.
Step 1: Use the Pythagorean theorem to find the vertical length. Let the longest side (hypotenuse) be mm and the horizontal length be mm. Let the vertical length be . The Pythagorean theorem states . The vertical length is approximately mm.
Step 2: Calculate the included angles using trigonometric ratios. Let be the angle opposite the vertical side () and adjacent to the horizontal side (). Let be the angle opposite the horizontal side () and adjacent to the vertical side (). Alternatively, since the sum of angles in a triangle is and one angle is : The vertical length is approximately mm, and the included angles are approximately and .
3. Work out the longest side of a right-angled triangle if the shorter sides are 50 mm and 220 mm.
Step 1: Use the Pythagorean theorem. Let the shorter sides be mm and mm. Let the longest side (hypotenuse) be . The longest side is approximately .
4. Work out the shorter side of a right-angled triangle if the longest side is 120 mm and the other shorter side is 90 mm.
Step 1: Use the Pythagorean theorem. Let the longest side (hypotenuse) be mm and one shorter side be mm. Let the other shorter side be . The shorter side is approximately .
5. Use your calculator to work out the ratios for these angles:
5.1 sin 20°
5.2 cos 30°
5.3 tan 45°
6. Use your calculator to work out the angles for these ratios:
6.1 sin x = 0.8337
6.2 cos x = 8/14
6.3 tan x = 32
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2. Calculate the vertical length and both included angles.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.