132k views
3 votes
The question is on the image, please help quickly!

The question is on the image, please help quickly!-example-1
User Rcgldr
by
8.8k points

1 Answer

5 votes

Answer:


c \approx \bf4748


\alpha \approx \bf 70.9^\circ


\beta = \bf 11.1^\circ

Explanation:

Use the attached diagram for help.

• To solve for c, we have to use the cosine rule, where:


c = √(a^2 + b^2 - 2ab \space\ cos \gamma)

Substituting the values:


c = √(4530^2 + 924^2 -2(4530)(924) * cos(98^\circ))


c \approx \bf4748

• Now that we know the value of c, we can use the sine rule to calculate the value of α (alpha):


(sin \space\ \alpha)/(a) = (sin \space\ \gamma)/(c)

Substituting the values:


(sin \space\ \alpha)/(4530) = (sin \space\ 98^\circ)/(4748)


sin \space\ \alpha = (sin \space\ 98^\circ)/(4748) * 4530


sin \space\ \alpha = 0.9448


\alpha \approx \bf 70.9^\circ

• Since we now have the values of both ∠α and ∠γ, we can find the value of ß using the triangle sum theorem:


\alpha + \beta + \gamma = 180^\circ


70.9^\circ + 98^\circ + \beta = 180^\circ


\beta + 168.9^\circ = 180^\circ


\beta = 180^\circ - 168.9^\circ


\beta = \bf 11.1^\circ

The question is on the image, please help quickly!-example-1
User JWL
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories