Question

Question 13

Solve ABC subject to the given conditions. Round the lengths of sides and measures of the angles to 1 decimal place if necessary

B= 28.1°, C= 57°, a=11/6

Answer 1

To solve for the remaining parts of triangle ABC, we can use the fact that the sum of the angles in a triangle is always 180 degrees. Therefore:

A = 180 - B - C

A = 180 - 28.1 - 57

A = 94.9 degrees

Now that we know all three angles of triangle ABC, we can use the Law of Sines or the Law of Cosines to find the lengths of the sides.

Using the Law of Sines, we have:

a / sin(A) = b / sin(B) = c / sin(C)

Let's use side b = 10 cm as a reference. Then:

a / sin(94.9) = 10 / sin(28.1)

a = sin(94.9) * 10 / sin(28.1)

a = 19.5 cm (rounded to 1 decimal place)

c / sin(57) = 10 / sin(28.1)

c = sin(57) * 10 / sin(28.1)

c = 16.6 cm (rounded to 1 decimal place)

Therefore, the sides of triangle ABC are approximately:

a = 19.5 cm

b = 10 cm

c =

Answer 2

∠A = 94.9°

b = 0.9 (1 d.p.)

c = 1.5 (1 d.p.)

Explanation:

Given conditions for triangle ABC:

• ∠B = 28.1°
• ∠C = 57°
• Side a = 11/6

To find the measure of the remaining angle, ∠A, we can use the Triangle Sum Theorem.

The Triangle Sum Theorem states that the interior angles of a triangle sum to 180°. Therefore:

`\begin{aligned}m \angle A + m \angle B+m \angle C&=180^(\circ)\\m \angle A + 28.1^(\circ)+57^(\circ)&=180^(\circ)\\m \angle A + 85.1^(\circ)&=180^(\circ)\\m \angle A + 85.1^(\circ)-85.1^(\circ)&=180^(\circ)-85.1^(\circ)\\m \angle A&=94.9^(\circ)\end{aligned}`

Now we have found the measure of angle A, we can use the Sine Rule to find the remaining side lengths, b and c.

`\boxed{\begin{minipage}{7.6 cm}\underline{Sine Rule} \\\\$(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)$\\\\\\where:\\ \phantom{ww}$\bullet$ $A, B$ and $C$ are the angles. \\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides opposite the angles.\\\end{minipage}}`

Substitute the given values into the formula:

`((11)/(6))/(\sin 94.9^(\circ))=(b)/(\sin 28.1^(\circ))=(c)/(\sin 57^(\circ))`

Therefore:

`(11)/(6\sin 94.9^(\circ))=(b)/(\sin 28.1^(\circ))=(c)/(\sin 57^(\circ))`

Solve for side b:

`\begin{aligned}(11)/(6\sin 94.9^(\circ))&=(b)/(\sin 28.1^(\circ))\\\\b&=(11\sin 28.1^(\circ))/(6\sin 94.9^(\circ))\\\\ b&=0.866689274...\\\\b&=0.9\; \sf (1\;d.p.)\end{aligned}`

Solve for side c:

`\begin{aligned}(11)/(6\sin 94.9^(\circ))&=(c)/(\sin 57^(\circ))\\\\c&=(11\sin 57^(\circ))/(6\sin 94.9^(\circ))\\\\ c&=1.54320265...\\\\c&=1.5 \; \sf (1\;d.p.)\end{aligned}`

Therefore, the angles and side lengths of triangle ABC, rounded to one decimal place where necessary, are:

• ∠A = 94.9°
• ∠B = 28.1°
• ∠C = 57°
• Side a = 11/6
• Side b = 0.9
• Side c = 1.5