Question

In ΔIJK, k = 7.7 cm, i = 3.4 cm and ∠J=30°. Find the length of j, to the nearest 10th of a centimeter.

Answer 1

its 5.1

Explanation:

\text{S.A.S.}\rightarrow \text{Law of Cosines}

S.A.S.→Law of Cosines

a^2=b^2+c^2-2bc\cos A

a

2

=b

2

+c

2

−2bccosA

From reference sheet.

j^2 = 7.7^2+3.4^2-2(7.7)(3.4)\cos 30

j

2

=7.7

2

+3.4

2

−2(7.7)(3.4)cos30

Plug in values.

j^2 = 59.29+11.56-2(7.7)(3.4)(0.866025)

j

2

=59.29+11.56−2(7.7)(3.4)(0.866025)

Square and find cosine.

j^2 = 59.29+11.56-45.34509

j

2

=59.29+11.56−45.34509

Multiply.

j^2 = 25.50491

j

2

=25.50491

Add.

j=\sqrt{25.50491} \approx5.05 \approx5.1

j=

25.50491

​

≈5.05≈5.1

Square root and round.

Answer 2

To the nearest tenth = 10.8cm

Explanation:

Using the cosine rule

J² = i² + k² + 2ikcosj

J² = 3.4² + 7.7² + 2(3.4)(7.7) cos 30

J² = 11.56 + 59.29 + 52.36cos30

J² = 11.56 + 59.29 + 52.36(0.8660)

J² = 11.56 + 59.29 + 45.35

J² = 116.2

J=√116.2

J= 10.78 cm

To the nearest tenth = 10.8cm