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Anon25 [30]
3 years ago
8

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 .

Mathematics
1 answer:
spayn [35]3 years ago
6 0

Answer:

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:

\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}

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

Using sine rule, we can find ∠K:

\frac{k}{sin(K)}=\frac{j}{sin(J)}   \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)}  \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o  \\

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