Some Applications of Trigonometry: CBSE Class 10 Mathematics practice questions

A. $\tan A$ or $\cot A$

$BC$ is opposite to $\angle A$ and $AB$ is adjacent, so $\tan A = \dfrac{BC}{AB}$ and $\cot A = \dfrac{AB}{BC}$. Sine and cosine need the hypotenuse $AC$, and so do secant and cosecant, so they do not link the two known sides.

B. $4\sqrt{3}$ m

The ladder is the hypotenuse and the height $h$ is opposite the $60^\circ$ angle, so $\sin 60^\circ = \dfrac{h}{8}$ and $h = 8 \cdot \dfrac{\sqrt 3}{2} = 4\sqrt3$ m. Using cosine gives $4$ m (the distance of the foot), tangent gives $8\sqrt{3}$ m, and cotangent gives $\dfrac{8\sqrt{3}}{3}$ m, each using the wrong sides.

A. $9\sqrt{3}$ m

Standing part: $\tan30^\circ=\dfrac{x}{9}$ gives $x=3\sqrt{3}$ m. Broken part (hypotenuse): $\cos30^\circ=\dfrac{9}{y}$ gives $y=6\sqrt{3}$ m. Original height $= x+y=9\sqrt{3}$ m. The options $3\sqrt{3}$ and $6\sqrt{3}$ give only one of the two parts, and 9 is just the distance on the ground.