Digital SAT Math : Right triangles and trigonometry - Practice Questions- New Syllabus
DSAT MAth Practice questions – all topics
- Geometry and Trigonometry Weightage: 15% Questions: 5-7
- Area and volume
- Lines, angles, and triangles
- Right triangles and trigonometry
- Circles
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DSAT MAth and English – full syllabus practice tests
▶️ Answer/Explanation
Ans: D
Since \( \triangle ABC \sim \triangle EDC \), the ratios of corresponding sides are equal.
Assume \( E \) is on \( AC \), \( D \) on \( BC \), with \( \angle ACB = \angle ECD = 90^\circ \).
Given \( AE = 15 \), let \( CE = 10 \), so \( AC = AE + CE = 25 \).
Similarity ratio \( \frac{AC}{EC} = \frac{25}{10} = \frac{5}{2} \), confirming \( CE = 10 \).
Question
In rectangle \(ABCD\) above \(E\) is on \bar{DC}, \(F\) is on \( \bar{BC} , DE = 6\) and \(FC =1\) .
If angle A is trisected (divided into three equal angles) by \(\bar{AE}\) and \(\bar{AF}, \) what is the length of \( \bar{BF}\)
▶️Answer/Explanation
Ans:
\(\angle DAB = 90^{\circ}\) is divided by 3 hence each angle = \(30^{\circ}\)
In triangle ADE
\(tan \theta =\frac{DE}{AD}\)
or
\(tan 30^{\circ} =\frac{6}{AD}\)
or
\(\frac{1}{\sqrt{3}}=\frac{6}{AD}\)
\(AD= 6\sqrt{3}\)
Now
\(BC = BF+FC\)
Hence
\(BF=BC-FC =AD-Fc as AD=BC\)
\(= 6\sqrt{3}-1\)
▶️ Answer/Explanation
Ans: B
\( \angle C \leftrightarrow \angle F \), so \( \cos F = \cos C \)
In \( \triangle DEF \), \( \angle E = 90^\circ \), sides 3k:4k:5k, \( \cos F = \frac{DF}{DE} = \frac{4}{5} \)
Choice A: Incorrect, uses \( \sin F \)
Choice C: Incorrect, inverts ratio
Choice D: Incorrect, uses \( \frac{\text{opposite}}{\text{adjacent}} \)