Suppose that m /_BCA=70^@, and x=33 cm and y = 47 cm. What is the degree measure of /_ABC?01510101651.png

Question

Suppose that $m ∠ B C A = 70^{\circ}$ , and $x = 33$ cm and $y = 47$ cm. What is the degree measure of $∠ A B C$ ?

Answer 1

A B = \sqrt{A C^{2} + B C^{2} - (2 \cdot A C \cdot B C \cos ∠ B C A)}

A B = \sqrt{47^{2} + 33^{2} - (2 \cdot 47 \cdot 33 {\cos 70}^{\circ})}

A B = \sqrt{2237.05}

A B = 47.29

Using the cosine law,

∠ A B C = \cos^{- 1} (\frac{B C^{2} + A B^{2} - A C^{2}}{2 \cdot B C \cdot A B})

∠ A B C = \cos^{- 1} (\frac{33^{2} + {47.29}^{2} - 47^{2}}{2 \cdot 33 \cdot 47.29})

∠ A B C = \cos^{- 1} (0.3576)

∠ A B C = \cos^{- 1} = {69.04}^{\circ}

Answer 2

Step 1

Given that

x=33 cm, y=47 cm

$m ∠ B C A = 70^{\circ}$

$∠ A B C = ?$

using cosine Rules

$\cos C = \frac{(A C)^{2} + (B C)^{2} - (A B)^{2}}{2 (A C) (B C)}$

$\cos 70^{\circ} = \frac{(47)^{2} + (33)^{2} - (A B)^{2}}{2 \times 47 \times 33}$

$0.3420 = \frac{3298 - (A B)^{2}}{3102}$

AB=47.30 cm

Using sine rule

$\frac{\sin ∠ A B C}{47} = \frac{\sin ∠ B C A}{47.30}$

$\sin ∠ A B C = \frac{47}{47.30} \times \sin 70^{\circ}$

$\sin ∠ A B C = 0.9337$

$∠ A B C \approx {68.24}^{\circ}$

$∠ A B C \approx {68.72}^{\circ}$

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