Question

Trigonometry question

Answer 1

4.59 cm to nearest hundredth.

Explanation:

Consider triangle ABD.

sin 72 = DB / 8

DB = 8 sin 72

= 7.608 cm.

Imagine a line from point C perpendicular to DB at point E.

As triangle CDB is isosceles ( DC = BC), this line will cut DB in 2

so, DE = 1/2 * 7.608 = 3.804.

Now consider triangle DCE which is a right-angled triangle at E.

cos 34 = 3.804 / DC

DC = 3.804 / cos 34

= 4.588 cm

BC = DC = 4.588.

Answer 2

`(8\sin 72^(\circ)\sin 34^(\circ))/(\sin 116^(\circ))`

Explanation:

In triangle ABD,

`\sin 72^(\circ)=(BD)/(8) \implies BD=8 \sin72^(\circ)`

In triangle BCD,

∠CBD=34° (isosceles triangle)

∠DCB=116° (sum of angles in a triangle)

`(BC)/(\sin 34^(\circ))=(8\sin 72^(\circ))/(\sin 116^(\circ)) \\ \\ BC=(8\sin 72^(\circ)\sin 34^(\circ))/(\sin 116^(\circ))`