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frika
You can calculate [tex]\sin22^{\circ}[/tex] using formula:
[tex]\cos^222^{\circ}+\sin^222^{\circ}=1[/tex].
Then [tex]\sin^222^{\circ}=1-\cos^222^{\circ}[/tex] and
[tex]\sin^222^{\circ}=1-( 0.9272)^2=0.14030016 \\ \sin22^{\circ}= \sqrt{0.14030016} =0.3746[/tex].
Answer: [tex]\sin22^{\circ}=0.3746[/tex]





Google817

Answer:

Step-by-step explanation:

In the given triangle ABC, ∠CAD = 22° segment CB = 15 cm.

We have to find sin of ∠C.

In the given triangle ADC,

∠CAD + ∠ADC + ∠DCA = 180

22° + 90° + ∠DCA = 180

∠DCA = 180 - 112 = 68°

Now we have to get the value of sinC or sin of angle DCA

Therefore sinC = sin 68° = 0.9272

Option B. 0.9272 is the correct option.

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