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What are the lengths of the missing sides of a 30-60-90 triangle if the longer leg of the triangle is 18 centimeter?
The longest side of the triangle will always be opposite the largest angle, which is 90° in this case. We can use the sine law to work out the other sides with that: sin(90°) / 18 = sin(60°) / x = sin(30°) / y 1/18 = sin(60°) / x x = 18 sin(60°) x = 18√3 / 2 x = 9√3 1/18 = sin(30°) / y y = 18 sin(30°) y = 9 So the triangle has a sides of 9 and 9√3, with a hypotenuse of 18
Cos 60 equals and sin 60 equals?
cos60= 1/2 sin60=1.732/2
What is the sin of 105 degrees?
sin 105 = sin (60+45) = sin60cos45 + cos60sin45sin 105 = ((sqrt(3)/2)((sqrt(2)/2)) + ((1/2)((sqrt(2)/2)))sin 105 = (sqrt(6) + sqrt(2)) / 4
What is the resultant of a 30 unit and 60 unit vector inclined to each other at 60 degrees?
Suppose the 30 unit vector is acting horizontally. Then the 60 unit vector has a horizontal component of 60*cos(60) units and a vertical component of 60*sin(60) units. So total horizontal = 30 + 60*cos(60) = 60 units total vertical = 60*sin(60)= 51.96 units. Then magnitude of resultant = sqrt(602 + 51.962) = sqrt(6300) = 79.37 units (approx). And direction = tan-1(51.96/60) = 40.89 degrees (from the 30 unit vector).
What is the area of an equilateral triangle when its perimeter is 22.5 cm?
Area = 0.5*7.52*sin(60 degrees) = 24.357 square cm rounded to 3 dp.
