Question

=7 \( 7^{2}-122 / \) \( \angle c= \) \[ \angle d=\frac{99^{\circ}}{} \] a) b) \( \left(\frac{0}{2}\right) \) \[ 1810^{2}-73^{3} \] \( =180^{\circ}-132^{\circ} \) \[ =\frac{108^{\circ}}{18} \times \] \[ =48^{2} V \] \( \angle b=48^{\circ}{ }^{\circ} \) C) \[ \begin{array}{l} \left.180^{\circ}-(90) \times 2\right) \\ = 18180^{\circ}-90^{\circ} \\ = 90^{\circ}-99 \\ = 180 \\ = 90^{\circ} \\ \\ \text { Review 3 } \end{array} \]

          =7 \( 7^{2}-122 / \)
\( \angle c= \)
\[
\angle d=\frac{99^{\circ}}{}
\]
a)
b)
\( \left(\frac{0}{2}\right) \)
\[
1810^{2}-73^{3}
\] \( =180^{\circ}-132^{\circ} \)
\[
=\frac{108^{\circ}}{18} \times
\]
\[
=48^{2} V
\]
\( \angle b=48^{\circ}{ }^{\circ} \)
C)
\[
\begin{array}{l}
\left.180^{\circ}-(90) \times 2\right) \\
= 18180^{\circ}-90^{\circ} \\
= 90^{\circ}-99 \\
= 180 \\
= 90^{\circ} \\
\\
\text { Review 3 }
\end{array}
\]

=7 7^2-122 /
∠ c=

∠ d=(99^∘)/()

a)
b)
((0)/(2))

1810^2-73^3
=180^∘-132^∘

=(108^∘)/(18)×

=48^2 V

∠ b=48^∘^∘
C)

.180^∘-(90) × 2)

= 18180^∘-90^∘

= 90^∘-99

= 180

= 90^∘

Review 3

Added by Gregg M.

Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Sanjiva Dayal IIT Kanpur 1st Edition

Chapter 13

Thanks for your feedback!

=7 \( 7^{2}-122 / \) \( \angle c= \) \[ \angle d=\frac{99^{\circ}}{} \] a) b) \( \left(\frac{0}{2}\right) \) \[ 1810^{2}-73^{3} \] \( =180^{\circ}-132^{\circ} \) \[ =\frac{108^{\circ}}{18} \times \] \[ =48^{2} V \] \( \angle b=48^{\circ}{ }^{\circ} \) C) \[ \begin{array}{l} \left.180^{\circ}-(90) \times 2\right) \\ = 18180^{\circ}-90^{\circ} \\ = 90^{\circ}-99 \\ = 180 \\ = 90^{\circ} \\ \\ \text { Review 3 } \end{array} \]