Right Triangle Trig word problems - Free Printable

Let's solve each problem step by step.

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1) Find the measure of angle $ \angle I $

We are given a right triangle with:
- One leg = 30
- Hypotenuse = 40
- Right angle at the bottom left
- We need to find $ m\angle I $, which is opposite the side of length 30.

Use sine because we have opposite and hypotenuse:

$$
\sin(\angle I) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{30}{40} = 0.75
$$

Now take the inverse sine (arcsin):

$$
\angle I = \sin^{-1}(0.75) \approx 48.6^\circ
$$

✔ Answer: $ m\angle I \approx 48.6^\circ $

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2) Find the measure of angle $ \angle J $

Given:
- One leg = 20
- Hypotenuse = 50
- Right angle at the bottom right
- Angle $ \angle J $ is adjacent to the 20 side and opposite the other leg.

Wait — let’s clarify the triangle.

From the diagram:
- The hypotenuse is 50
- One leg is 20 (adjacent to $ \angle J $)
- So $ \angle J $ is adjacent to the 20 side, and the hypotenuse is 50

So use cosine:

$$
\cos(\angle J) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{20}{50} = 0.4
$$

$$
\angle J = \cos^{-1}(0.4) \approx 66.4^\circ
$$

✔ Answer: $ m\angle J \approx 66.4^\circ $

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3) Building height from angle of elevation

Given:
- Distance from observer to building base = 78 ft
- Angle of elevation = $ 57^\circ $
- We want the height $ h $ of the building

This forms a right triangle where:
- Adjacent side = 78 ft
- Opposite side = $ h $
- Angle = $ 57^\circ $

Use tangent:

$$
\tan(57^\circ) = \frac{h}{78}
$$

$$
h = 78 \cdot \tan(57^\circ)
$$

Calculate $ \tan(57^\circ) \approx 1.5399 $

$$
h \approx 78 \cdot 1.5399 \approx 119.91
$$

Round to nearest tenth: $ \boxed{119.9} $ feet

✔ Answer: The building is approximately 119.9 feet high.

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4) Height of flagpole from shadow and angle

Given:
- Shadow length = 49 ft
- Angle of sun’s rays = $ 38^\circ $
- Want height $ x $ of flagpole

Right triangle:
- Adjacent = 49 ft
- Opposite = $ x $
- Angle = $ 38^\circ $

Use tangent:

$$
\tan(38^\circ) = \frac{x}{49}
$$

$$
x = 49 \cdot \tan(38^\circ)
$$

$ \tan(38^\circ) \approx 0.7813 $

$$
x \approx 49 \cdot 0.7813 \approx 38.28
$$

Round to nearest tenth: $ \boxed{38.3} $ feet

✔ Answer: The flagpole is approximately 38.3 feet tall.

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✔ Final Answers:

1) $ m\angle I \approx \boxed{48.6^\circ} $
2) $ m\angle J \approx \boxed{66.4^\circ} $
3) Building height = $ \boxed{119.9} $ feet
4) Flagpole height = $ \boxed{38.3} $ feet

Let me know if you'd like diagrams or more explanation!