Geometry and Trigonometry

QUESTION 1 [2 marks] Easy

A right triangle has legs of length 6 and 8. Find the length of the hypotenuse.

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By the Pythagorean theorem: $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$.

QUESTION 2 [2 marks] Easy

In a right triangle, one leg has length 5 and the hypotenuse has length 13. Find the length of the other leg.

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By the Pythagorean theorem: leg $=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$.

QUESTION 3 [3 marks] Medium

In right triangle $ABC$, the right angle is at $C$, $\angle A=30^\circ$, and the hypotenuse $AB=10$. Find the length of side $BC$ (opposite $\angle A$).

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$BC=AB\cdot\sin(A)=10\cdot\sin(30^\circ)=10\times0.5=5$.

QUESTION 4 [3 marks] Medium

If $\sin(\theta)=\dfrac{3}{5}$ and $\theta$ is an acute angle, find $\cos(\theta)$.

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A $3$-$4$-$5$ right triangle has $\sin(\theta)=\dfrac{3}{5}$ (opposite/hypotenuse), so the adjacent side is $\sqrt{5^2-3^2}=\sqrt{16}=4$. Thus $\cos(\theta)=\dfrac{4}{5}$.

QUESTION 5 [4 marks] Hard

A 13-foot ladder leans against a wall so that the base of the ladder is 5 feet from the wall. To the nearest degree, what angle does the ladder make with the ground?

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The ground, wall, and ladder form a right triangle with hypotenuse 13 (the ladder) and one leg 5 (the base distance, adjacent to the angle with the ground). So $\cos(\theta)=\dfrac{5}{13}\approx0.3846$, giving $\theta=\cos^{-1}(0.3846)\approx67^\circ$.

QUESTION 6 [5 marks] Hard

In right triangle $XYZ$, the right angle is at $Y$. If $\tan(X)=\dfrac{5}{12}$, what is $\sin(X)$?

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$\tan(X)=\dfrac{5}{12}$ means the side opposite $X$ and the side adjacent to $X$ are in ratio $5:12$. Since $5,12,13$ is a Pythagorean triple ($5^2+12^2=25+144=169=13^2$), the hypotenuse is 13. So $\sin(X)=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{5}{13}$.

QUESTION 7 [5 marks] Hard

A surveyor stands 40 meters from the base of a building and measures the angle of elevation to the top as $35°$. Find the height of the building, to the nearest tenth of a meter.

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The height, the horizontal distance (40 m), and the line of sight form a right triangle where the height is opposite the $35°$ angle and 40 m is adjacent to it. So $\text{height}=40\tan(35°)\approx40(0.7002)\approx28.0$ meters.

Right triangles and trigonometry 7 questions