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3 years ago

In triangle ABC, AB = 6 cm, BC = 13 cm and angle ACB = 23°.

Mathematics

1 answer:

amid [387]3 years ago

5 0

Answer:

angle BAC = 57.84°

Step-by-step explanation:

To find angle BAC given AB=6cm BC= 13 cm and ACB=23°

we will follow the steps below;

we can use the sine formula to solve the problem given

$\frac{sin A}{a}$ = $\frac{sin C}{c}$

from the question given;

A = ?

C = 23°

c=AB = 6 cm

a=BC = 13 cm

substitute the values into the formula above

$\frac{sin A}{a}$ = $\frac{sin C}{c}$

$\frac{sin A}{13}$ = $\frac{sin 23}{6}$

cross -multiply

6 sinA = 13 sin23°

sin A = 13 sin23° / 6

sin A = 0.84658

take the $sin^{-1}$ of both-side

$sin^{-1}$ sin A = $sin^{-1}$ (0.84658)

on the left-hand side of the equation $sin^{-1}$ will cancel-out sin leaving us with just A

A = $sin^{-1}$ (0.84658)

A ≈ 57.84°

angle BAC = 57.84°

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Hello from MrBillDoesMath!

Answer:

Discussion:

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Check the answer:

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Does 34 + 2(3) = 40 ?

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