In \Delta ABC, side a = 5\text{ cm}, side b = 5\sqrt{2}\text{ cm}, and \angle A = 30^\circ… — JAMB UTME

In $\Delta ABC$ , side $a = 5\text{ cm}$ , side $b = 5\sqrt{2}\text{ cm}$ , and $\angle A = 30^\circ$ . What are all possible values for $\angle B$ ?

45^\circ

135^\circ

Applying the Sine Rule gives

\frac{5}{\sin 30^\circ} = \frac{5\sqrt{2}}{\sin B}

, so

\sin B = \frac{\sqrt{2}}{2}

. The angles whose sine is

\frac{\sqrt{2}}{2}

between

0^\circ

and

180^\circ

are

45^\circ

and

135^\circ

. Checking angle sums:

30^\circ + 45^\circ = 75^\circ < 180^\circ

and

30^\circ + 135^\circ = 165^\circ < 180^\circ

, so both

45^\circ

and

135^\circ

yield valid triangles.

Set up the Sine Rule formula relating sides

a

b

and angles

A

B

\frac{a}{\sin A} = \frac{b}{\sin B}

The Sine Rule connects two sides and their opposite angles in any non-right triangle.

Substitute given values

a = 5

b = 5\sqrt{2}

, and

A = 30^\circ

\frac{5}{\sin 30^\circ} = \frac{5\sqrt{2}}{\sin B} \implies \frac{5}{0.5} = \frac{5\sqrt{2}}{\sin B} \implies 10 = \frac{5\sqrt{2}}{\sin B}

Since

\sin 30^\circ = \frac{1}{2}

, simplifying yields the ratio.

Solve for

\sin B

\sin B = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2}

Isolating

\sin B

gives the principal trigonometric ratio value.

Determine all valid values for angle

B

in the range

(0^\circ, 180^\circ)

Acute

B_1 = \arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

; Obtuse

B_2 = 180^\circ - 45^\circ = 135^\circ

. Both are valid since

30^\circ + 135^\circ = 165^\circ < 180^\circ

Since

b > a

, the ambiguous case (SSA) produces two valid distinct triangle solutions.

Sine Rule and the Ambiguous Case (SSA)

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