A ramp for a wheelchair has an angle of elevation of 5°. If the ramp is 12 feet long, how high off the ground is the top of the ramp? (Given sin 5° ≈ 0.087)
Answer: B
Let h be the height and L be the length of the ramp.
\(\sin(\text{angle}) = \frac{\text{height}}{\text{length}}\)
\(\sin 5° = \frac{h}{12}\)
h = 12 × \(\sin 5°\) ≈ 12 × 0.087 = 1.044 feet.
Please log in or register to report this question.
If \(A+B=90°\), then \(\sqrt{\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}}\) is equal to:
Answer: A
Since A+B=90°, we have B=90-A. So, \(\tan B = \cot A\), \(\cot B = \tan A\), \(\sec B = \csc A\), \(\sin B = \cos A\).
Substitute these into the expression under the root:
Numerator: \(\tan A \cot A + \tan A \tan A = 1 + \tan^2 A = \sec^2 A\).
Denominator: \(\sin A \csc A = 1\).
Second term: \(\frac{\cos^2 A}{\cos^2 A} = 1\).
The expression becomes \(\sqrt{\frac{\sec^2 A}{1} - 1} = \sqrt{\sec^2 A - 1}\).
Using the identity \(\sec^2 A - 1 = \tan^2 A\), we get \(\sqrt{\tan^2 A} = \tan A\).
Please log in or register to report this question.
The value of \(\tan 315°\) is:
Answer: B
315° lies in the fourth quadrant, where the tangent is negative.
\[ \tan 315^{\circ} = -\tan(360^{\circ} - 315^{\circ}) = -\tan 45^{\circ} = -1 \]
Please log in or register to report this question.
What is the value of \(\sin(-60°)\)?
Answer: B
Sine is an odd function.
\[ \sin(-\theta) = -\sin\theta \]
\[ \sin(-60^{\circ}) = -\frac{\sqrt{3}}{2} \]
Cosine, by contrast, is even: \( \cos(-\theta) = \cos\theta \).
Please log in or register to report this question.
The minimum value of \(\sec^2\theta + \csc^2\theta\) is:
Answer: C
We rewrite the expression using identities:
\(\sec^2\theta + \csc^2\theta = (1+\tan^2\theta) + (1+\cot^2\theta) = 2 + \tan^2\theta + \cot^2\theta\).
Let \(y = \tan^2\theta\). The expression is \(2 + y + 1/y\). By AM-GM inequality, \(y+1/y \ge 2\sqrt{y \cdot 1/y} = 2\). The minimum value of \(y+1/y\) is 2.
Therefore, the minimum value of the entire expression is 2 + 2 = 4.
Please log in or register to report this question.
The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. The tangent of one of its acute angles is:
Answer: D
Let the two legs of the right triangle be a and b. Hypotenuse c = 25 cm.
Perimeter = a + b + c = 60. So, a + b + 25 = 60 => a + b = 35.
By Pythagorean theorem, a² + b² = c² = 25² = 625.
We know \((a+b)^2 = a^2+b^2+2ab\). So, \(35^2 = 625 + 2ab \Rightarrow 1225 = 625 + 2ab \Rightarrow 2ab = 600 \Rightarrow ab = 300\).
We need two numbers that add to 35 and multiply to 300. The numbers are 15 and 20.
The legs are 15 cm and 20 cm. Let one acute angle be A. The tangent values would be opposite/adjacent, which could be 15/20 = 3/4 or 20/15 = 4/3.
Please log in or register to report this question.
The value of \(\sin^2 1° + \sin^2 2° + ... + \sin^2 89°\) is:
Answer: C
We pair the terms using the identity \(\sin(90-A) = \cos A\).
The series is \((\sin^2 1° + \sin^2 89°) + (\sin^2 2° + \sin^2 88°) + ... + \sin^2 45°\).
Each pair \((\sin^2 A + \sin^2(90-A))\) becomes \((\sin^2 A + \cos^2 A) = 1\).
There are 44 such pairs (from 1° to 44°). So the sum of these pairs is 44.
The middle term left over is \(\sin^2 45° = (1/\sqrt{2})^2 = 1/2 = 0.5\).
Total sum = 44 + 0.5 = 44.5.
Please log in or register to report this question.
If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is:
Answer: B
Let h be the height of the tower and s be the length of the shadow. The angle of elevation θ is given by \(\tan \theta = h/s\).
Since the height h is constant, if the shadow length 's' increases, the value of the fraction h/s decreases.
As \(\tan \theta\) decreases, the angle θ also decreases (for acute angles). Therefore, the angle of elevation is decreasing.
Please log in or register to report this question.
