Hey
I suppose it will be like
A vector into B vector is (3*12)+(4*5)=36+20= 56
AB= root under (3 square+4 square) * root under (12 sqaure + 5 square)= 5+13=18
Hence mod of cos theta will be : 56/18= 3.11(which is not possible hence I guess you got the question wrong and it should have been either -4j cap or -5 j cap(one of these)).
Question : If $\triangle A B C \sim \triangle F D E$ such that $A B=9 \mathrm{~cm}, A C=11 \mathrm{~cm}, D F=16 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$, then the length of $BC$ is:
Option 1: $5 \frac{3}{4} \mathrm{~cm}$
Option 2: $4\frac{3}{5} \mathrm{~cm}$
Option 3: $3\frac{5}{7} \mathrm{~cm}$
Option 4: $6\frac{3}{4} \mathrm{~cm}$
Question : $\triangle \mathrm{ABC}$ is a right-angle triangle at $\mathrm{B}$. If $\tan \mathrm{A}=\frac{5}{12}$, then $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}$ will be equal to:
Option 1: $1 \frac{5}{13}$
Option 2: $2 \frac{4}{13}$
Option 3: $3 \frac{1}{13}$
Option 4: $2 \frac{1}{13}$
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