(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.Given $AC = 16$ (hypotenuse) and $BC = 8$ (side adjacent to $\angle C$).Using the trigonometric ratio for

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(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.Given $AC = 16$ (hypotenuse) and $BC = 8$ (side adjacent to $\angle C$).Using the trigonometric ratio for cosine:$\cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{8}{16} = \frac{1}{2}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $m \angle C = 60^{\circ}$.

Class 10 Mathematics Introduction to Trigonometry Trigonometric Ratios of Some Specific Angles

Medium

In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 16$ and $BC = 8$,then $m \angle C = \dots$ (in $^{\circ}$)

A
$30$
B
$45$
C
$60$
D
$90$

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Class 10

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Chapter

Introduction to Trigonometry

Subtopic

Trigonometric Ratios of Some Specific Angles

If $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$,then $x = \ldots$ (in $^{\circ}$)

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In $\triangle PQR$,right-angled at $Q$,$PQ = 3 \, cm$ and $PR = 6 \, cm$. Determine $\angle QPR$ and $\angle PRQ$.

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In $\triangle ABC$,right-angled at $B$,$AB = 5 \, cm$ and $\angle ACB = 30^{\circ}$. Determine the lengths of the sides $BC$ and $AC$.

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If $\cos \theta = \frac{1}{\sqrt{2}},$ then $\theta = \ldots$ (in $^\circ$)

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A = 60^{\circ}$. If $AC = 10$,then $AB = \dots$

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In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 16$ and $BC = 8$,then $m \angle C = \dots$ (in $^{\circ}$)

Similar Questions

If $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$,then $x = \ldots$ (in $^{\circ}$)

In $\triangle PQR$,right-angled at $Q$,$PQ = 3 \, cm$ and $PR = 6 \, cm$. Determine $\angle QPR$ and $\angle PRQ$.

In $\triangle ABC$,right-angled at $B$,$AB = 5 \, cm$ and $\angle ACB = 30^{\circ}$. Determine the lengths of the sides $BC$ and $AC$.

If $\cos \theta = \frac{1}{\sqrt{2}},$ then $\theta = \ldots$ (in $^\circ$)

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A = 60^{\circ}$. If $AC = 10$,then $AB = \dots$

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