Grade 10 · Geometry & measurement

Similar right triangle trigonometry practice

Similar right triangle trigonometry is a grade 10 math skill aligned to Common Core standard HSG.SRT.C.6: understand that side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 similar right triangle trigonometry problems our math games drill.

CCSS HSG.SRT.C.610 questions in the bank
Sample questions

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Question 1easy

In right triangle ABCABC, C\angle C is the right angle. The three sides measure AB=61AB = 61, BC=60BC = 60, and CA=11CA = 11. Triangle ABCABC is similar to triangle XYZXYZ, where BB corresponds to YY and CC corresponds to ZZ. What is sinY\sin Y?

Because BB corresponds to YY, sinY=sinB\sin Y = \sin B. In right triangle ABCABC, sinB=oppositehypotenuse=CAAB=1161\sin B = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{CA}{AB} = \dfrac{11}{61}.

Question 2easy

In right triangle ABCABC, C\angle C is the right angle. The three sides measure AB=37AB = 37, BC=35BC = 35, and CA=12CA = 12. Triangle ABCABC is similar to triangle XYZXYZ, where BB corresponds to YY and CC corresponds to ZZ. What is cosY\cos Y?

Because BB corresponds to YY, cosY=cosB\cos Y = \cos B. In right triangle ABCABC, cosB=adjacenthypotenuse=BCAB=3537\cos B = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{BC}{AB} = \dfrac{35}{37}.

Question 3medium

In right triangle ABCABC, C\angle C is the right angle. The three sides measure AB=25AB = 25, BC=24BC = 24, and CA=7CA = 7. Triangle ABCABC is similar to triangle DEFDEF, where AA corresponds to DD and CC corresponds to FF. What is tanD\tan D?

Because AA corresponds to DD, tanD=tanA\tan D = \tan A. In right triangle ABCABC, tanA=oppositeadjacent=BCCA=247\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{BC}{CA} = \dfrac{24}{7}.

Question 4medium

In right triangle ABCABC, A\angle A is the right angle. The three sides measure BC=15BC = 15, CA=12CA = 12, and AB=9AB = 9. Triangle ABCABC is similar to triangle XYZXYZ, where BB corresponds to YY and AA corresponds to XX. What is sinY\sin Y?

Because BB corresponds to YY, sinY=sinB\sin Y = \sin B. With the right angle at AA, sinB=oppositehypotenuse=CABC=1215=45\sin B = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{CA}{BC} = \dfrac{12}{15} = \dfrac{4}{5}.

Question 5medium

In right triangle ABCABC, C\angle C is the right angle. The three sides measure AB=29AB = 29, BC=21BC = 21, and CA=20CA = 20. Triangle ABCABC is similar to triangle XYZXYZ, where AA corresponds to XX and CC corresponds to ZZ. What is cosX\cos X?

Because AA corresponds to XX, cosX=cosA\cos X = \cos A. In right triangle ABCABC, cosA=adjacenthypotenuse=CAAB=2029\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{CA}{AB} = \dfrac{20}{29}.

Question 6medium

In right triangle ABCABC, C\angle C is the right angle. The three sides measure AB=41AB = 41, BC=40BC = 40, and CA=9CA = 9. Triangle ABCABC is similar to triangle DEFDEF, where BB corresponds to EE and CC corresponds to FF. What is tanE\tan E?

Because BB corresponds to EE, tanE=tanB\tan E = \tan B. In right triangle ABCABC, tanB=oppositeadjacent=CABC=940\tan B = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{CA}{BC} = \dfrac{9}{40}.

Question 7medium

In right triangle ABCABC, B\angle B is the right angle. The three sides measure AC=13AC = 13, BC=12BC = 12, and AB=5AB = 5. Triangle ABCABC is similar to triangle XYZXYZ, where AA corresponds to XX and BB corresponds to YY. What is sinX\sin X?

Because AA corresponds to XX, sinX=sinA\sin X = \sin A. With the right angle at BB, sinA=oppositehypotenuse=BCAC=1213\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{12}{13}.

Question 8medium

In right triangle ABCABC, A\angle A is the right angle. The three sides measure BC=17BC = 17, CA=15CA = 15, and AB=8AB = 8. Triangle ABCABC is similar to triangle DEFDEF, where CC corresponds to FF and AA corresponds to DD. What is tanF\tan F?

Because CC corresponds to FF, tanF=tanC\tan F = \tan C. With the right angle at AA, tanC=oppositeadjacent=ABCA=815\tan C = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{AB}{CA} = \dfrac{8}{15}.

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