In JKL, k=4.1 cm, j=3.8 cm and angle J=Q3^ Find all possible values of angle K , to the nearest inch of a degree .

74.0°

Step-by-step explanation:

In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the  nearest 10th of a degree

Solution:

A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

then, the surface area of the sphere shown in the image is: 256π sq. units!

step-by-step explanation:

the lateral surface area of a sphere is 2πrh where h=2r.

then:

surface area = 4πr^2.

in this case, r=8 units ⇒ surface area = 4π(8 units)^2. = 256π sq. units!

then, the surface area of the sphere shown in the image is: 256π sq. units!