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Mathematics · 2024 · 2 marks
CBSE 2024 · Region 1 · Set 1 · Q23
(A) Evaluate: 2 √ 2 cos 45^° sin 30^°+2 √ 3 cos 30^°
(A)
Evaluate : $\displaystyle 2 \sqrt{2} \cos 45^{\circ} \sin 30^{\circ}+2 \sqrt{3} \cos 30^{\circ}$
OR
(B)
If $\displaystyle \mathrm{A}=60^{\circ}$ and $\displaystyle \mathrm{B}=30^{\circ}$, verify that : $\displaystyle \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
Marking-scheme solution
\[\begin{aligned}
& 2 \sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}+2 \sqrt{3} \times \frac{\sqrt{3}}{2} \\
& =4
\end{aligned}
\]
\[\begin{aligned}
& \text { LHS }=\sin \left(60^{\circ}+30^{\circ}\right)=\sin 90^{\circ}=1 \\
& \text { RHS }=\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ} \\
& \quad=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}=1 \\
& \therefore \text { LHS }=\text { RHS }
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.