SL 3.3—Angles of elevation and depression, bearings

The observer is 45 m from the base (adjacent side), the angle of elevation is 50°, and the height $h$ is the unknown (opposite side).

Since we know the adjacent side and want the opposite side, we use: $\tan (50°)=\frac{h}{45}$

$h=45\times \tan (50°)\approx 45\times 1.1918\approx 53.6\text{ m}$

The height of the flagpole is approximately 53.6 m (3 s.f.).

We need the opposite side of a right-angled triangle where the adjacent is 45 m and the angle is 50°. So $h=45\tan (50°)$.

34.5 m corresponds to $45\times \tan (37.5°)$ or using $\sin (50°)$ incorrectly with the wrong side. Since $\tan (50°)>1$, the height must exceed 45 m, so 34.5 m is too small.

29.0 m is approximately $45\times \tan (33°)$, which corresponds to a much smaller angle than 50°. This is eliminated.

69.9 m would require $\tan (\theta )\approx 1.55$, giving $\theta \approx 57°$, not 50°. This arises from using $\frac{45}{\cos (50°)}$ (the hypotenuse), not the height.

The only consistent answer is 53.6 m, confirming $45\times \tan (50°)\approx 53.6$ m.

The string is the hypotenuse (80 m), the angle of elevation is 55°, and the height $h$ is the opposite side.

Since opposite and hypotenuse are involved: $\sin (55°)=\frac{h}{80}$

$h=80\times \sin (55°)\approx 80\times 0.8192\approx 65.5\text{ m}$

The kite is approximately 65.5 m above the ground.

The string is the hypotenuse (80 m) and the height is the opposite side, so we need $h=80\sin (55°)$.

45.9 m $\approx 80\times \cos (55°)$, which gives the horizontal distance, not the height. This is the adjacent side, not the opposite.

98.0 m $\approx 80\times \tan (55°)\approx 80\times 1.428$. This incorrectly treats the horizontal distance as 80 m instead of the string length.

114 m exceeds the string length of 80 m, which is impossible for the height of a kite on an 80 m string.

65.5 m is correct: $80\times \sin (55°)\approx 65.5$ m.