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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 5 · Set 1 · Q24
Evaluate: $\displaystyle \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}$For acute angles A and B and A + 2B and 2A + B are acute if $\displaystyle \tan (\mathrm{A}+2 \mathrm{~B})=\sqrt{3}$ and $\displaystyle \sin (2 \mathrm{~A}+\mathrm{B})=\frac{1}{\sqrt{2}}$, then find the measures of angles A and B.
Evaluate: $\displaystyle \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}$
For acute angles A and B and A + 2B and 2A + B are acute if $\displaystyle \tan (\mathrm{A}+2 \mathrm{~B})=\sqrt{3}$ and $\displaystyle \sin (2 \mathrm{~A}+\mathrm{B})=\frac{1}{\sqrt{2}}$, then find the measures of angles A and B.
Marking-scheme solution
\[\begin{aligned}
& \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}=\frac{\left(\dfrac{\sqrt{3}}{2}\right)^{3}-\dfrac{1}{\sqrt{3}}}{\left(\dfrac{1}{\sqrt{2}}\right)^{2}} \\
& =\frac{1}{4 \sqrt{3}} \text { or } \frac{\sqrt{3}}{12}
\end{aligned}
\]
(b)
\[\begin{aligned}
& \tan (A+2 B)=\sqrt{3} \Rightarrow A+2 B=60^{\circ} \\
& \sin (2 A+B)=\frac{1}{\sqrt{2}} \Rightarrow 2 A+B=45^{\circ}
\end{aligned}
\]
On solving above equations, \(\displaystyle \mathrm{A}=10^{\circ}, \mathrm{B}=25^{\circ}\)
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.