Consider the non-right triangle below Suppose that m∠ACB=98∘ and m∠BAC=42∘, and that y=51.4cm. What is the value of x?

Step 1 Solution: in △ABC we know that ∠A+∠B+∠C=180∘ Now in △ABC (given) ∠BAC+∠ACB+∠ABC=180∘ 42∘+98∘+∠ABC=180∘ ∠ABC=180∘−140∘ ∠ABC=40∘ Using Sine angle rule asin⁡A=bsin⁡B=csin⁡c Step 2 xsin⁡∠BAC=ysin⁡∠ABC=ABsin⁡∠ACB xsin⁡42∘=51.4sin⁡40∘=ABsin⁡98∘ taking first turo ratio x0.67=5.40.64 ⇒x=51.4×0.670.64 ⇒x=53.73cm

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High School
Geometry
Trigonometry
Non-right triangles and trigonometry

danrussekme

Answered question

2021-11-22

Consider the non-right triangle below

Suppose that

m ∠ A C B = 98^{\circ}

and

m ∠ B A C = 42^{\circ}

, and that

y = 51.4 c m

. What is the value of x?

Answer & Explanation

Mary Darby

Beginner2021-11-23Added 11 answers

Step 1
Solution: in

△ A B C

we know that

∠ A + ∠ B + ∠ C = 180^{\circ}

Now in

△ A B C

(given)

∠ B A C + ∠ A C B + ∠ A B C = 180^{\circ}

42^{\circ} + 98^{\circ} + ∠ A B C = 180^{\circ}

∠ A B C = 180^{\circ} - 140^{\circ}

∠ A B C = 40^{\circ}

Using Sine angle rule

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin c}

Step 2

\frac{x}{\sin ∠ B A C} = \frac{y}{\sin ∠ A B C} = \frac{A B}{\sin ∠ A C B}

\frac{x}{{\sin 42}^{\circ}} = \frac{51.4}{{\sin 40}^{\circ}} = \frac{A B}{{\sin 98}^{\circ}}

taking first turo ratio

\frac{x}{0.67} = \frac{5.4}{0.64}

\Rightarrow x = \frac{51.4 \times 0.67}{0.64}

\Rightarrow x = 53.73 c m

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