# Law of Sines Worksheet

Law of Sines Worksheet
• Page 1
1.
In Δ$\mathrm{ABC}$ if $\angle$$A$ = 35o, $b$ = 15, and $c$ = 28, then find the area of triangle $\mathrm{ABC}$ to three significant digits.

 a. 420 b. 210 c. 172 d. 120

#### Solution:

Area of triangle ABC = 12 bc sin A

= 12(15)(28) sin 35o = 120
[Simplify using calculator.]

2.
In ΔABC, if $\angle$B = 48o, $a$ = 12 and $c$ = 6, then find the area of triangle ABC to two significant digits.

 a. 24 b. 27 c. 36 d. 72

#### Solution:

Area of triangle ABC = 12ac sin B

= 12(12)(6) sin 48o = 27
[Simplify using calculator.]

3.
In ΔPQR, if $\angle$R = 68o, $p$ = 14 and $q$ = 8, then find the area of triangle PQR to two significant digits.

 a. 56 b. 21 c. 42 d. 52

#### Solution:

Area of triangle PQR = 12 pq sin R

= 12(14)(8) sin 68o = 52

4.
In ΔABC, if $\angle$A = 54o, $\angle$B = 62o, and $b$ = 10, then the length of $a$ to two significant digits is:

 a. 8.9 units b. 11 units c. 9.2 units d. 10 units

#### Solution:

Sin 54oa = Sin 62o10
[Using law of sines: sin Aa = sin Bb.]

a = 10 Sin 54oSin 62o

a = 9.2, to two significant digits.

5.
In ΔABC, if $\angle$A = 62o, $b$ = 12, and $a$ = 20, then what is the measure of the $\angle$B to the nearest degree?

 a. 32o b. 148o c. 59o d. 16o

#### Solution:

sin 62o20 = sin B12
[Using law of Sines : sin Aa = sin Bb .]

sin B = 12 sin 62o20

sin B = 0.529768555...
[Simplify using calculator.]

B = 32o (or) B = 148o

The measure of 148o is not possible.
[Since 148o + 62o = 210o and 210o is greater than 180o.]

So, B = 32o.

6.
In ΔDEF, if $\angle$F = 80o, $f$ = 12, $e$ = 26, then the measure of angle E to the nearest degree is:

 a. 27o b. 63o c. 80o d. Cannot be determined

#### Solution:

sin E26 = sin 80o12
[Using law of sines: sin Ee = sin Ff .]

sin E = 26 sin 80o12

sin E = 2.133750132
[Simplify using calculator.]

This is not possible, because the sine of an angle can be no greater than one.

Therefore, a triangle with the given measurements is not possible.

7.
In ΔDEF, if $\angle$D = 43.4o, $d$ = 10.8 cm and $e$ = 6.4 cm, then what is the area of the triangle to three significant digits?

 a. 34.56 cm2 b. 31.9 cm2 c. 27.05 cm2 d. 12.16 cm2

#### Solution:

sin 43.4o10.8 = sin E6.4
[Using law of sines: sin Dd = sin Ee .]

sin E = 6.4 sin 43.4o10.8

sin E = 0.407162969

E = 24o

F = 180o - (24o + 43.4o) = 112.6o

Area of triangle DEF = 12 de sin F

= 12(10.8)(6.4) sin 112.6o
[Simplify using calculator.]

= 31.9 cm2

8.
In ΔABC, if $\angle$E = 34o20', $\angle$D = 68o30' and $f$ = 16.5 in., then the length of $d$ to three significant digits is:
 a. 16.5 units b. 15.2 units c. 17.3 units d. 15.7 units

#### Solution:

F = 180o - (34o20' + 68o30')

F = 77o10'
[Sum of angle measures in a triangle is 180o.]

sin 77o10′16.5 = sin 68o30′d
[Use the law of sines: sin Ff = sin Dd.]

d = 16.5 sin 68o30′sin 77o10′

d = 15.7 units
[Simplify using calculator.]

9.
In ΔABC, if $\angle$B = 58o24', $b$ = 23.6 cm and $a$ = 8.2 cm, then the measure of the angle A to the nearest tenth of one degree is:

 a. 72.8o b. 8o24′ c. 60o8′ d. 17.2o

#### Solution:

sin A8.2 = sin 58o24′23.6
[Use law of sines: sin Aa = sin Bb.]

sin A = 8.2 Sin 58o24′23.6

sin A = 0.295939019
[Simplify using calculator.]

A = 17.2o

10.
In quadrilateral ABCD, if $\angle$BAD = 90o, AB = 3 units, AD = 4 units, then the length of BC to the nearest two significant digits is:

 a. 4.3 units b. 5 units c. 3.5 units d. 5.8 units

#### Solution:

BD2 = 32 + 42 = 25

BD = 5

In Triangle BDC, using law of sines, sin C / BD = sin ∠BDCBC .

sin 60o5 = sin ∠BDCBC

BC = 5sin 48osin 60o

BC = 4.3

The length of BC is 4.3 units.