Q1:2026Geometry > TrigonometryEasyIf a pole is 6 meters high and casts a shadow $2\sqrt{3}$ meters long on the ground, then the sun's elevation isAnswer options$45^{\circ}$$30^{\circ}$$90^{\circ}$$60^{\circ}$Correct AnswerOption 4Correct AnswerExplanation →
Q2:2026Geometry > TrigonometryMediumIf $\cos\alpha + \cos\beta + \cos\gamma = 3$, then $\cos^3\alpha + \cos^3\beta + \cos^3\gamma$ will beAnswer options2130Correct AnswerOption 3Correct AnswerExplanation →
Q3:2026Geometry > TrigonometryMediumIn $\triangle ABC$, $\angle A = 90^{\circ}$, if $\tan C = \sqrt{3}$, then find the value of $\sin B + \cos C - \cos^2 B$Answer options1/21/401/8Correct AnswerOption 2Correct AnswerExplanation →
Q4:BMSAT Kozhikode 2025Geometry > TrigonometryMediumIf $1 + \sin^2\theta = 3 \sin \theta \cos \theta$, then $\tan \theta$ can take valuesAnswer options$1,\frac{1}{2}$$1,2$$\frac{1}{2},2$None of theseCorrect AnswerOption 1Correct AnswerExplanation →
Q5:BMSAT Kozhikode 2025Geometry > TrigonometryEasyIf a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then the sun's elevation isAnswer options60°45°30°None of theseCorrect AnswerOption 1Correct AnswerExplanation →
Q6:BMSAT Kozhikode 2025Geometry > TrigonometryEasy$\dfrac{\sin \theta}{(1+ \cos \theta)}$ is equal toAnswer options$\frac{(1+ \cos \theta)}{\sin \theta}$$\frac{(1-\cos \theta)}{\cos \theta}$$\frac{(1-\cos \theta)}{\sin \theta}$$\frac{(1-\sin \theta)}{\cos \theta}$Correct AnswerOption 3Correct AnswerExplanation →