12 multiple-choice questions, progressively harder.
A ray at an obtuse angle meets the unit semicircle at height yyy, where 0<y<10<y<10<y<1. Which expression is its signed horizontal projection?
Solution
Correct answer: D
The unit radius and its projections satisfy the Pythagorean theorem. The horizontal coordinate must take the negative root because the angle is obtuse.
Triangle P has sides of lengths sss and ttt meeting at 57∘57^\circ57∘. Triangle Q has sides of lengths 3s3s3s and 12t\tfrac12t21t meeting at 123∘123^\circ123∘. How does the area of Q compare with the area of P?
Correct answer: B
The angles have the same sine by reflection. In the area formula, the side product is multiplied by 333 and then halved.
A triangular banner has area 48.5 ft248.5\text{ ft}^248.5 ft2. One side at its 109∘109^\circ109∘ corner measures 11.311.311.3 ft. Find the length of the other side at that corner to the nearest tenth of a foot.
Correct answer: C
Let uuu be the unknown side length. Solve the area equation for uuu, and evaluate in degree mode.
Two rigid rods are joined at a hinge. The acute setting 36∘36^\circ36∘ gives a target triangle area. Which other setting strictly between 0∘0^\circ0∘ and 180∘180^\circ180∘ gives exactly the same area?
Correct answer: A
With the rod lengths fixed, the area fixes the height on the unit semicircle. A horizontal line at that height meets the semicircle in the original point and its reflection in the vertical axis.
In triangle ABCABCABC, angle AAA is acute and ABABAB is horizontal to the right. The length of ACACAC is bbb, and the length of ABABAB is ccc. When does the perpendicular from CCC to line ABABAB land strictly beyond BBB?
The foot is at horizontal coordinate bcosAb\cos AbcosA measured from AAA. Point BBB is at horizontal coordinate ccc. The foot lies strictly beyond BBB when its coordinate is greater.
In triangle ABCABCABC, let ABABAB be the horizontal base of length ccc and let DDD be the perpendicular foot from CCC. Angle AAA is obtuse and CDCDCD has length hhh. The larger triangle BCDBCDBCD contains ABCABCABC and ACDACDACD. Which subtraction gives the area of ABCABCABC?
From left to right the points are DDD, AAA, and BBB. The larger triangle has base AD+cAD+cAD+c. Subtract the triangle on base ADADAD. Both heights are hhh.
The included angle of a triangle increases from 98∘98^\circ98∘ to 166∘166^\circ166∘, while the two side lengths at the angle stay fixed. Why does the area decrease?
The upper-left arc descends from the top toward the left endpoint. The sine stays positive but decreases. The base and the length used to scale the height stay fixed.
One triangle has two sides of lengths 141414 and 252525 meeting at angle TTT. Another triangle has sides of lengths 101010 and 353535 meeting at 180∘−T180^\circ-T180∘−T. Which comparison is correct?
The included angles have equal sines. The two side products both equal 350350350. The products therefore give equal areas even though the pairs of lengths differ.
A triangle has area 117 m2117\text{ m}^2117 m2 and two sides of lengths 151515 m and 202020 m. Which equation determines the sine of the included angle θ\thetaθ?
Use the given sides at the included angle. Divide the area by half their product.
Two points on the upper unit semicircle have the same nonzero horizontal coordinate. What can you conclude about their angles between 0∘0^\circ0∘ and 180∘180^\circ180∘?
A vertical line through a fixed horizontal coordinate meets the upper semicircle at just one point. The rays therefore end at the same point and make the same angle. The nonnegative vertical coordinate is fixed by the radius equation.
A triangular sail has two edges of lengths 12.212.212.2 m and 9.19.19.1 m. The angle between them is 63∘63^\circ63∘. Which estimate of the area to the nearest square meter uses the correct calculation?
Multiply half the product of the edges by the sine of their included angle. Round the final result to the nearest integer.
The area of a triangle is exactly half the product of two of its side lengths. What must be true of the angle between those two sides?
The area formula already contains half the product. Both side lengths are positive, so equality requires the sine to be 111. This occurs at the top of the unit semicircle, at 90∘90^\circ90∘.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.