Trigonometric Ratios in Real-World Problems

Solution

The building is upright on level ground, so $\angle B = 90^{\circ}$.

From the $33^{\circ}$ angle at $A$: the height $BT$ is opposite and $AB = 38$ m is adjacent. Opposite and adjacent: tangent.

$\tan 33^{\circ} = \dfrac{h}{38}$, so $h = 38 \tan 33^{\circ} = 24.677\ldots = 24.7$ (3 s.f.)

Solution

The tower is vertical, so $\angle F = 90^{\circ}$. From the $37^{\circ}$ angle at $A$: $TF = 40$ m is opposite and $FA$ is adjacent: tangent.

$\tan 37^{\circ} = \dfrac{40}{d}$, so $d = \dfrac{40}{\tan 37^{\circ}} = 53.081\ldots = 53.1$ (3 s.f.)

Solution

Each position of the ladder is the hypotenuse of a right-angled triangle with the wall, and the height reached is opposite the angle with the ground: sine.

Top edge: $WT = 5 \sin 78^{\circ} = 4.890\ldots$ m

Lower edge: $WL = 5 \sin 62^{\circ} = 4.414\ldots$ m

$TL = WT - WL = 4.890\ldots - 4.414\ldots = 0.4760\ldots$ m $= 47.60\ldots$ cm $= 47.6$ cm (3 s.f.)

Solution

From the $5^{\circ}$ angle at $F$: the rise $0.5$ m is opposite, the ramp is the hypotenuse and $FB$ is adjacent.

(a) $\sin 5^{\circ} = \dfrac{0.5}{l}$, so $l = \dfrac{0.5}{\sin 5^{\circ}} = 5.736\ldots = 5.74$ (3 s.f.)

(b) $\tan 5^{\circ} = \dfrac{0.5}{d}$, so $d = \dfrac{0.5}{\tan 5^{\circ}} = 5.715\ldots = 5.72$ (3 s.f.)

A gentle slope needs a long ramp: at $5^{\circ}$ it is more than eleven times as long as it is high.

Solution

The line $AM$ from the top to the midpoint of the base is perpendicular to it, so $\triangle ABM$ has a right angle at $M$ and $BM = 10 \div 2 = 5$ m.

From the $30^{\circ}$ angle at $B$: $BM = 5$ m is adjacent, $AB$ is the hypotenuse and $AM$ is opposite.

$\cos 30^{\circ} = \dfrac{5}{AB}$, so $AB = \dfrac{5}{\cos 30^{\circ}} = 5.773\ldots = 5.77$ m (3 s.f.)

$\tan 30^{\circ} = \dfrac{AM}{5}$, so $AM = 5 \tan 30^{\circ} = 2.886\ldots = 2.89$ m (3 s.f.)

Solution

The pole is vertical, so $\angle P = 90^{\circ}$. The ray of sunlight that just passes the top $T$ ends at the tip $S$ of the shadow.

From $\theta$ at $S$: the pole $6$ m is opposite and the shadow $4$ m is adjacent: tangent.

$\theta = \tan^{-1}\left(\dfrac{6}{4}\right) = 56.309\ldots = 56.3^{\circ}$ (1 d.p.)

Solution

Both buildings are upright and $PH$ is horizontal, so $\angle PHS = \angle PHR = 90^{\circ}\text{,}$ and $HS = PQ = 32$ m.

In $\triangle PHS$: from $\angle HPS$, $HS = 32$ m is opposite and $PH$ adjacent: $PH = \dfrac{32}{\tan 22.4^{\circ}} = 77.637\ldots$ m

In $\triangle PHR$: from $\angle RPH$, $RH$ is opposite and $PH$ adjacent: $RH = 77.637\ldots \times \tan 40.7^{\circ} = 66.779\ldots$ m

$RS = RH + HS = 66.779\ldots + 32 = 98.779\ldots = 98.8$ m (3 s.f.)

With $PH$ rounded to $77.6$ m, the answer would come out as $98.7$ m: keep the full value.