Question

EASY TRIG - 15 POINTS

Answer 1

Answer: The answers are (a) 40 cm and (b) $\sin^{-1}\dfrac{23}{62}.$

Step-by-step explanation: The calculations are as follows:

(a) See the figure (a). As given in the question, A circle with centre 'O' circumscribes a triangle ABC with BC = 20 cm and ∠BAC = 30°. We need to find the diameter DC of the circle.

Let us draw BD. Now, ∠BAC and ∠BDC are angles on the same arc BC, so we have

∠BAC = ∠BDC = 30°.

Also, ∠CBD = 90°, since it stands on the diameter DC. So, ΔBCD will be a right angled triangle.

We can write

$\sin \angle BDC=\dfrac{BC}{DC}\\\\\\ \Rightarrow \sin 30^\circ=\dfrac{20}{DC}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{20}{DC}\\\\\\\Rightarrow DC=40.$

Thus, the diameter of the circle = 40 cm.

(b) See the figure (b).

As given in the question, A circle with centre 'O'' circumscribes a triangle DEF with EF = 4.6 inches and diameter GF = 12.4 in.. We need to find the angle EDF.

Let us draw GE. Now, ∠EGF and ∠EDF are angles on the same arc EF, so we have

∠EGF = ∠EDF = ?

Also, ∠GEF = 90°, since it stands on the diameter GF. So, ΔGEF will be a right angled triangle.

We can write

$\sin \angle EGF=\dfrac{EF}{GF}\\\\\\ \Rightarrow \sin \angle EGF=\dfrac{4.6}{12.4}\\\\\\\Rightarrow \sin \angle EGF=\dfrac{23}{62}\\\\\\\Rightarrow \angle EGF=\sin^{-1}\dfrac{23}{62}.$

Thus,

$\angle EDF=\angle EGF=\sin^{-1}\dfrac{23}{62}.$