14.3°
55.9°
59.5°
75.6°
Correct answer is C
→r.→t=|→r||→t|cosθ
→r.→t=(3i+4j).(−5i+12j)=−15+48=33
|→r|=√32+42=√25=5
|\overrightarrow{t}| = \sqrt{(-5)^{2} + 12^{2}| = \sqrt{169} = 13
cosθ=→r.→t|→r||→t|
cosθ=335×13=3365
\theta = \cos^{-1} {\frac{33}{65}} \approxeq 59.5°
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