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Last year exam question of bca 1 sem?
sin60+cos135 /sin60*cos135
The hypotenuse of a 30-60-90 triangle has a length of 15 what is the length of the side opposite the 60 angle?
The length of the side opposite the 60° angle is: 12.99 units.The long leg is (sin60°)h = 0.866 h = 12.99
What is the hypotenuse of a 30-60-90 triangle if the long leg is 4?
Sin60=4/H where H is the length of the hypotenuse. So H = 4/sin(60) = 4/[sqrt(3)/2] = 8/sqrt(3) = 4.62 units (approx)
What is the area of a parallelogram with sides of 12 cm and 16 cm and an included angle of 60 degrees?
To find the area you would first have to find the height of the parallelogram. Height: Sin60=x/12cm .866=x/12cm 10.4cm=x With the height you can then find the area Area = Base * Height Area=16cm*10.4cm Area=166.3cm^2
In a 30-60-90 right triangle the length of the long leg is 5 what is the length of the hypotenuse is?
use Pythagorean theorem. A squared+ B squared= C squared. the you square root C squared.from superscot85: ...which is fine if you know two of the sides, which you don't. The side opposite the 60o angle is 5, so the hypotenuse is 5/sin60 ie 5.77. (Now you can use Pythagoras to get the short side: sqrt(5.772 - 52) ie sqrt 8.293 which is 2.88.)
