This geometry right triangle worksheet answer key solves all 22 problems of the right triangle trigonometry worksheet step by step.
This answer key gives a full worked solution for every problem on the right triangle trigonometry worksheet. Compare each setup, not just the final number, and check the rounding rules. Calculators must be in degree mode.
Geometry Right Triangle Worksheet Answer Key: problems 1–5
Problem 1: Name the sides
The side across from \(\angle A\) is \(\overline{BC}\). The leg that touches \(\angle A\) (other than the hypotenuse) is \(\overline{AB}\). The side across from the right angle is \(\overline{AC}\).
Opposite: \(\overline{BC}\); adjacent: \(\overline{AB}\); hypotenuse: \(\overline{AC}\).
Problem 2: Ratios from side lengths
Relative to \(A\), the opposite side is 8, the adjacent side is 15 and the hypotenuse is 17.
\(\sin A=\frac{8}{17}\), \(\cos A=\frac{15}{17}\), \(\tan A=\frac{8}{15}\).
\(\sin A=\frac{8}{17},\ \cos A=\frac{15}{17},\ \tan A=\frac{8}{15}\)
Problem 3: Ratios for the other angle
The side opposite \(D\) is \(EF=21\). The side opposite \(E\) is \(DF=20\).
\(\sin D=\frac{21}{29}\), \(\cos E=\frac{21}{29}\), \(\tan E=\frac{20}{21}\).
\(\sin D=\cos E=\frac{21}{29}\). They are equal because \(D\) and \(E\) are complementary.
Problem 4: Ratios in a 7-24-25 triangle
The side opposite \(P\) is \(QR=7\) and the side adjacent to \(P\) is \(PQ=24\).
\(\tan P=\frac{7}{24}\), \(\cos P=\frac{24}{25}\). The side opposite \(R\) is \(PQ=24\), so \(\sin R=\frac{24}{25}\).
\(\tan P=\frac{7}{24},\ \cos P=\frac{24}{25},\ \sin R=\frac{24}{25}\)
Problem 5: Using a calculator
Enter each angle with the calculator in degree mode.
\(\sin 28^{\circ}\approx 0.4695\), \(\cos 63^{\circ}\approx 0.4540\), \(\tan 51^{\circ}\approx 1.2349\).
\(0.4695,\ 0.4540,\ 1.2349\)
Geometry Right Triangle Worksheet Answer Key: problems 6–10
Problem 6: Finding an angle from a ratio
Use the inverse function for each ratio.
\(x=\sin^{-1}(0.6)\approx 36.9^{\circ}\); \(x=\cos^{-1}(0.25)\approx 75.5^{\circ}\); \(x=\tan^{-1}(2.4)\approx 67.4^{\circ}\).
\(36.9^{\circ},\ 75.5^{\circ},\ 67.4^{\circ}\)
Problem 7: Similar right triangles
In triangle ABC, \(\sin A=\frac{9}{15}=\frac{3}{5}\). Since \(\angle D\cong\angle A\), \(\sin D=\sin A=\frac{3}{5}\). In triangle DEF, \(\sin D=\frac{EF}{DE}=\frac{21}{DE}\). Set the ratios equal: \(\frac{21}{DE}=\frac{3}{5}\), so \(3\cdot DE=105\) and \(DE=35\).
Check: \(\frac{21}{35}=\frac{3}{5}\).
\(DE=35\)
Problem 8: Complementary angle identities
Subtract each angle from \(90^{\circ}\) and switch between sine and cosine.
\(\sin 20^{\circ}=\cos 70^{\circ}\), \(\cos 75^{\circ}=\sin 15^{\circ}\), \(\sin 38^{\circ}=\cos 52^{\circ}\).
\(\cos 70^{\circ},\ \sin 15^{\circ},\ \cos 52^{\circ}\)
Problem 9: Solving with complementary angles
Sine and cosine are equal only for complementary angles, so the angles add to \(90^{\circ}\).
\((3x+5)+(2x+10)=90\), so \(5x+15=90\), \(5x=75\) and \(x=15\).
Check: \(\sin 50^{\circ}=\cos 40^{\circ}\).
\(x=15\)
Problem 10: 45-45-90 triangle, find the hypotenuse
In a 45-45-90 triangle the hypotenuse is the leg times \(\sqrt{2}\).
\(x=7\sqrt{2}\approx 9.9\).
\(x=7\sqrt{2}\approx 9.9\)
Geometry Right Triangle Worksheet Answer Key: problems 11–15
Problem 11: 45-45-90 triangle, find the legs
The legs equal the hypotenuse divided by \(\sqrt{2}\).
\(\text{leg}=\frac{12}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{12\sqrt{2}}{2}=6\sqrt{2}\approx 8.5\).
\(6\sqrt{2}\approx 8.5\text{ in}\)
Problem 12: 30-60-90 triangle, short leg given
The side opposite \(30^{\circ}\) is the short leg, 5. The hypotenuse is twice the short leg, and the long leg is the short leg times \(\sqrt{3}\).
\(x=2(5)=10\) and \(y=5\sqrt{3}\approx 8.7\).
\(x=10,\ y=5\sqrt{3}\approx 8.7\)
Problem 13: 30-60-90 triangle, hypotenuse given
The short leg is half the hypotenuse, and the long leg is the short leg times \(\sqrt{3}\).
\(\text{short leg}=\frac{18}{2}=9\) and \(\text{long leg}=9\sqrt{3}\approx 15.6\).
\(9\text{ cm},\ 9\sqrt{3}\text{ cm}\)
Problem 14: 30-60-90 triangle, long leg given
The long leg equals the short leg times \(\sqrt{3}\), so the short leg is \(\frac{6\sqrt{3}}{\sqrt{3}}=6\). The hypotenuse is twice the short leg: \(2(6)=12\).
\(\text{short leg }6\text{ ft},\ \text{hypotenuse }12\text{ ft}\)
Problem 15: Angle from two sides
The known sides are opposite and adjacent, so use tangent: \(\tan A=\frac{7}{11}\).
\(A=\tan^{-1}(\frac{7}{11})\approx 32.5^{\circ}\).
\(m\angle A\approx 32.5^{\circ}\)
Geometry Right Triangle Worksheet Answer Key: problems 16–20
Problem 16: Solve a right triangle
The acute angles are complementary: \(\angle B=90^{\circ}-34^{\circ}=56^{\circ}\).
Side \(a\) is opposite \(A\): \(\sin 34^{\circ}=\frac{a}{20}\), so \(a=20\sin 34^{\circ}\approx 11.2\).
Side \(b\) is adjacent: \(\cos 34^{\circ}=\frac{b}{20}\), so \(b=20\cos 34^{\circ}\approx 16.6\).
Check: \(11.2^{2}+16.6^{2}\approx 401.0\approx 20^{2}\).
\(\angle B=56^{\circ},\ a\approx 11.2,\ b\approx 16.6\)
Problem 17: Ladder against a wall
The height is opposite the \(70^{\circ}\) angle and the ladder is the hypotenuse, so use sine.
\(\sin 70^{\circ}=\frac{h}{15}\), so \(h=15\sin 70^{\circ}\approx 14.1\).
\(h\approx 14.1\text{ ft}\)
Problem 18: Height of a flagpole
The distance is adjacent to the angle and the part of the pole above her eyes is opposite, so use tangent.
\(\tan 32^{\circ}=\frac{h}{85}\), so \(h=85\tan 32^{\circ}\approx 53.1\) ft above eye level.
Add her eye height: \(53.1+5.5=58.6\).
\(\approx 58.6\text{ ft}\)
Problem 19: Lighthouse and boat
The angle of depression equals the angle of elevation from the boat, so the angle at the boat is \(25^{\circ}\). The 120 ft height is opposite and the distance \(d\) is adjacent.
\(\tan 25^{\circ}=\frac{120}{d}\), so \(d=\frac{120}{\tan 25^{\circ}}\approx 257.3\).
\(d\approx 257.3\text{ ft}\)
Problem 20: Wheelchair ramp
The rise is opposite the angle. The ramp surface is the hypotenuse and the horizontal distance is adjacent.
Length: \(\sin 4.8^{\circ}=\frac{2.5}{L}\), so \(L=\frac{2.5}{\sin 4.8^{\circ}}\approx 29.9\).
Horizontal distance: \(\tan 4.8^{\circ}=\frac{2.5}{x}\), so \(x=\frac{2.5}{\tan 4.8^{\circ}}\approx 29.8\).
\(\text{ramp}\approx 29.9\text{ ft},\ \text{horizontal}\approx 29.8\text{ ft}\)
Geometry Right Triangle Worksheet Answer Key: problems 21–22
Problem 21: Drone height
The 85 ft path is the hypotenuse and the rise is opposite the \(33^{\circ}\) angle, so use sine.
\(85\sin 33^{\circ}\approx 46.3\) ft of rise. Add the 6 ft platform: \(46.3+6=52.3\).
\(\approx 52.3\text{ ft}\)
Problem 22: Flagpole on a building
Find each height with tangent, then subtract.
To the roof: \(40\tan 50^{\circ}\approx 47.7\). To the top of the pole: \(40\tan 58^{\circ}\approx 64.0\).
Flagpole: \(64.0-47.7\approx 16.3\), which agrees with \(40(\tan 58^{\circ}-\tan 50^{\circ})\approx 16.3\).
\(\approx 16.3\text{ ft}\)
How to use this geometry right triangle worksheet answer key
Read this geometry right triangle worksheet answer key after trying each problem yourself, and redo any problem you missed.
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