Solve 13

Solve the missing dimensions for the following triangle: Triangle ABC: AngleA=43 degrees, b=7.0cm, c=6.0cm

Question 1. Angle B with units written as degrees
Question 2. Angle C with units written as degrees
Question 3. a, rounded to the nearest tenth of a cm with the unit.

Correct answer:

a = 4.8545 cm
B = 79.5497 °
C = 57.4503 °

Step-by-step explanation:

A = 43^{\circ} b = 7.0 cm c = 6.0 cm a^{2} = b^{2} + c^{2} - 2 b c cos A a = b^{2} + c^{2} - 2 \cdot b \cdot c \cdot cos A = b^{2} + c^{2} - 2 \cdot b \cdot c \cdot cos 43 ° = 7^{2} + 6^{2} - 2 \cdot 7 \cdot 6 \cdot cos 43 ° = 7^{2} + 6^{2} - 2 \cdot 7 \cdot 6 \cdot 0.731354 = 4.855 = 4.8545 cm

sin B : sin A = b : a = 7 : 4.8545 ≐ 1.442 = 82653 : 57320 b_{0} = sin A \cdot \frac{b}{a} = sin 43 ° \cdot \frac{b}{a} = sin 43 ° \cdot \frac{7}{4 . 8 5 4 5} = 0.681998 \cdot \frac{7}{4 . 8 5 4 5} = 0.98341 B = \frac{1 8 0 °}{π} \cdot arcsin (b_{0}) = \frac{1 8 0 °}{π} \cdot arcsin 0.9834 = 79.549 7^{\circ} = 79 ° 3 2^{'} 59 "

sin C : sin A = c : a = 6 : 4.8545 ≐ 1.236 = 68968 : 55801 c_{0} = sin A \cdot \frac{c}{a} = sin 43 ° \cdot \frac{c}{a} = sin 43 ° \cdot \frac{6}{4 . 8 5 4 5} = 0.681998 \cdot \frac{6}{4 . 8 5 4 5} = 0.84293 C = \frac{1 8 0 °}{π} \cdot arcsin (c_{0}) = \frac{1 8 0 °}{π} \cdot arcsin 0.8429 ≐ 57.450 3^{\circ} Verifying Solution: s = A + B + C = 43 + 79.5497 + 57.4503 = 180^{\circ}

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