Given a right triangle, where one cathetus is A―=11 m, the hypothenuse is C―=15 m, and ∠c=90∘ . Find all missing sides (B―) and angles (∠aand∠b).

first, to find B―, use the Pythagorean Theorem: A―2+B―2=C―2 Isolate B―2 B―2=C―2−A―2 B―2=152−112 B―=104 B―≈10.2 m To find ∠a, using the cosine ratio (assuming A― is adj to agn≤a): cos⁡∠a=A―C―=1115 ∠a=cos⁡(−1)(1115) ≈43∘ To find ∠b, subtract both other angles from 180∘ (since the sum of all angles in any triangle is 180∘): ∠b=180∘−∠a−∠c ∠b=180∘−43∘−90∘ ∠b=47∘ Answer: B―≈10.2m.∠a≈43∘.∠b≈47∘

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rocedwrp

Answered question

2021-09-06

Given a right triangle, where one cathetus is $\overset{―}{A} = 11$ m, the hypothenuse is $\overset{―}{C} = 15$ m, and $∠ c = 90^{\circ}$ . Find all missing sides $(\overset{―}{B})$ and angles $(∠ a and ∠ b)$ .

Answer & Explanation

davonliefI

Skilled2021-09-07Added 79 answers

first, to find

\overset{―}{B}

, use the Pythagorean Theorem:

{\overset{―}{A}}^{2} + {\overset{―}{B}}^{2} = {\overset{―}{C}}^{2}

Isolate

{\overset{―}{B}}^{2}

{\overset{―}{B}}^{2} = {\overset{―}{C}}^{2} - {\overset{―}{A}}^{2}

{\overset{―}{B}}^{2} = 15^{2} - 11^{2}

\overset{―}{B} = \sqrt{104}

\overset{―}{B} \approx 10.2

m
To find

∠ a

, using the cosine ratio (assuming

\overset{―}{A}

is adj to

a g n \leq a

\cos ∠ a = \frac{\overset{―}{A}}{\overset{―}{C}} = \frac{11}{15}

∠ a = \cos (- 1) (\frac{11}{15})

\approx 43^{\circ}

To find

∠ b

, subtract both other angles from

180^{\circ}

(since the sum of all angles in any triangle is

180^{\circ}

∠ b = 180^{\circ} - ∠ a - ∠ c

∠ b = 180^{\circ} - 43^{\circ} - 90^{\circ}

∠ b = 47^{\circ}

Answer:

\overset{―}{B} \approx 10.2 m . ∠ a \approx 43^{\circ} . ∠ b \approx 47^{\circ}

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